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\ihead{The $q$-Poincar\'e sum}\ohead{page \thepage}
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\begin{document}

\title{On the $q$-Poincar\'{e} sum of a Coxeter group}
\date{draft, \today}
\author{Benjamin Adenbaum, Darij Grinberg}
\maketitle

\begin{abstract}
\textbf{Abstract.} Given a finite Coxeter group $W$ and an element $q$ of a
commutative ring $\mathbf{k}$, we define the element $\mathbf{L}_{q}%
:=\sum_{w\in W}q^{\ell\left(  w\right)  }w$ of its group algebra
$\mathbf{k}\left[  W\right]  $. We say that an element $w\in W$ is
$W$\emph{-length-symmetric} if its coefficient in the commutator $\left[
\mathbf{L}_{a},\mathbf{L}_{b}\right]  $ is $0$ for all $\mathbf{k}$ and all
$a,b\in\mathbf{k}$. We prove several sufficient criteria for elements to be
$W$-length-symmetric. In particular, we show that any separable permutation
(i.e., permutation avoiding the patterns $2413$ and $3142$) in the symmetric
group $S_{n}$ is $S_{n}$-length-symmetric. This is only a sufficient
condition; other $W$-length-symmetric elements include (twisted) involutions
and elements of dihedral parabolic subgroups.

\end{abstract}

Fix a commutative ring $\mathbf{k}$. Consider the $n$-th symmetric group
$S_{n}$ for a given nonnegative integer $n$. Any permutation $w\in S_{n}$ has
a \emph{length} $\ell\left(  w\right)  $, which is defined as the number of
inversions of $w$ (that is, of pairs $\left(  i,j\right)  $ of integers with
$1\leq i<j\leq n$ and $w\left(  i\right)  >w\left(  j\right)  $). Using these
lengths, we can define an element%
\[
\mathbf{L}_{q}:=\sum_{w\in S_{n}}q^{\ell\left(  w\right)  }w\in\mathbf{k}%
\left[  S_{n}\right]
\]
in the group algebra $\mathbf{k}\left[  S_{n}\right]  $ of $S_{n}$, for any
fixed scalar $q\in\mathbf{k}$. For instance, $\mathbf{L}_{1}$ and
$\mathbf{L}_{-1}$ are two central elements of $\mathbf{k}\left[  S_{n}\right]
$, equalling $n!$ times the central idempotents corresponding to the trivial
and sign representations of $S_{n}$. For general $q$, however, the element
$\mathbf{L}_{q}$ is not central, and does not even necessarily commute with
another such element $\mathbf{L}_{r}$ for $r\neq q$.

However, it turns out that if we fix two scalars $q,r\in\mathbf{k}$, then many
permutations $w\in S_{n}$ do not appear in the commutator $\left[
\mathbf{L}_{q},\mathbf{L}_{r}\right]  :=\mathbf{L}_{q}\mathbf{L}%
_{r}-\mathbf{L}_{r}\mathbf{L}_{q}$. Namely, if $w\in S_{n}$ is a
\emph{separable} permutation, then the coefficient of $w$ in $\left[
\mathbf{L}_{q},\mathbf{L}_{r}\right]  $ is $0$. The notion of a separable
permutation can be defined in many ways, e.g., algorithmically or by the
avoidance of two patterns ($2413$ and $3142$); we refer to Ghys's book
\cite[\textquotedblleft Separable permutations\textquotedblright]{Ghys16} for
four equivalent definitions and a survey of their rich theory.

In this paper, we will prove the fact just mentioned, and essentially
generalize it to arbitrary finite Coxeter groups.

\section{The theorems}

We assume the basic properties of Coxeter groups (see \cite{BjoBre},
\cite{Davis}, \cite{Lusztig}, \cite{Bourbaki-Lie4-6}, \cite{Heckman}). A
Coxeter system $\left(  W,S\right)  $ consists of a Coxeter group $W$ and its
set $S$ of simple reflections. If $\left(  W,S\right)  $ is a Coxeter system,
and $w\in W$ is any element, then $\ell\left(  w\right)  $ denotes the Coxeter
length of $w$ (that is, the smallest length of a list of simple reflections
having product $w$). Note that $\ell\left(  w^{-1}\right)  =\ell\left(
w\right)  $ for each $w\in W$.

If $W$ is a finite Coxeter group, then there exists a unique element $w\in W$
whose length is maximum (see \cite[\S 11.1]{Lusztig}); we shall call it
$w_{0}$. This element $w_{0}$ is always an involution (i.e., satisfies
$w_{0}^{2}=\operatorname*{id}$), since its inverse $w_{0}^{-1}$ must also have
maximum length. It is known as the \emph{longest element} or the \emph{longest
word} of $W$.

We now define our main object of study:

\begin{definition}
\label{def.Lq}Let $\left(  W,S\right)  $ be a Coxeter system with finite
Coxeter group $W$. Let $\mathbf{k}$ be a commutative ring. For any
$q\in\mathbf{k}$, we define the element%
\[
\mathbf{L}_{q}\left(  W\right)  :=\sum_{w\in W}q^{\ell\left(  w\right)  }w
\]
in the group algebra $\mathbf{k}\left[  W\right]  $. We denote this element
simply by $\mathbf{L}_{q}$ if the group $W$ is clear from the context. We call
it the $q$\emph{-Poincar\'{e} sum} of $W$.
\end{definition}

\begin{example}
\label{exa.LqSn}Let $n$ be a positive integer. The symmetric group $S_{n}$ is
the Coxeter group of type $A_{n-1}$; the corresponding Coxeter system is
$\left(  S_{n},S\right)  $, where $S=\left\{  s_{1},s_{2},\ldots
,s_{n-1}\right\}  $ is the set of all simple reflections in $S_{n}$ (so that
$s_{i}$ is the transposition swapping $i$ with $i+1$). The length $\ell\left(
w\right)  $ of a permutation $w\in S_{n}$ is well-known to equal the number of
inversions of $w$ (that is, of pairs $\left(  i,j\right)  $ of elements of
$\left\{  1,2,\ldots,n\right\}  $ satisfying $i<j$ and $w\left(  i\right)
>w\left(  j\right)  $). The $q$-Poincar\'{e} sum $\mathbf{L}_{q}\left(
S_{n}\right)  =\sum_{w\in S_{n}}q^{\ell\left(  w\right)  }w\in\mathbf{k}%
\left[  S_{n}\right]  $ is then a linear combination of all permutations of
$\left\{  1,2,\ldots,n\right\}  $, where each $w$ appears with coefficient
$q^{\ell\left(  w\right)  }$. For instance,%
\[
\mathbf{L}_{q}\left(  S_{3}\right)  =\operatorname*{id}+\,qs_{1}+qs_{2}%
+q^{2}s_{1}s_{2}+q^{2}s_{2}s_{1}+q^{3}w_{0},
\]
where $w_{0}=s_{1}s_{2}s_{1}=s_{2}s_{1}s_{2}$ is the permutation sending
$1,2,3$ to $3,2,1$.

If $q$ is a nonnegative real, then the element $\mathbf{L}_{q}\left(
S_{n}\right)  $ can be viewed as a measure on $S_{n}$ (by treating each
coefficient $q^{\ell\left(  w\right)  }$, divided by the total sum of these
coefficients, as the measure of $w\in S_{n}$). This measure is known as the
\emph{Mallows measure} $\operatorname*{Mallows}\left(  n,q\right)  $ (see,
e.g., \cite{Korotk16}).
\end{example}

\begin{example}
Let $\left(  W,S\right)  $ be any Coxeter system with finite Coxeter group
$W$. Then,%
\[
\mathbf{L}_{1}\left(  W\right)  =\sum_{w\in W}w\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \mathbf{L}_{-1}\left(  W\right)  =\sum_{w\in
W}\underbrace{\left(  -1\right)  ^{\ell\left(  w\right)  }}%
_{=\operatorname*{sign}w}w
\]
are two elements in the center of $\mathbf{k}\left[  W\right]  $, and in fact
(up to a factor of $\left\vert W\right\vert $) are the central idempotents
corresponding to the trivial and sign representations of $W$, respectively. Of
course, $\mathbf{L}_{0}\left(  W\right)  =1$ lies in the center of
$\mathbf{k}\left[  W\right]  $, too.
\end{example}

However, in general, $\mathbf{L}_{q}=\mathbf{L}_{q}\left(  W\right)  $ does
not lie in the center of $\mathbf{k}\left[  W\right]  $ when $q\notin\left\{
1,-1,0\right\}  $; and in fact, the elements $\mathbf{L}_{q}$ for two
different values of $q$ usually do not commute. That is, most finite Coxeter
groups $W$ do \textbf{not} satisfy the equality $\mathbf{L}_{a}\mathbf{L}%
_{b}=\mathbf{L}_{b}\mathbf{L}_{a}$ when $a$ and $b$ are generic.

This note is devoted to salvaging this equality to some extent. First, we
shall show that it holds when $W$ is dihedral (or smaller):

\begin{theorem}
\label{thm.comm.dihe}Let $\left(  W,S\right)  $ be a Coxeter system, where $W$
is finite and $\left\vert S\right\vert \leq2$. Then,
\[
\mathbf{L}_{a}\mathbf{L}_{b}=\mathbf{L}_{b}\mathbf{L}_{a}%
\]
for any $a,b\in\mathbf{k}$.
\end{theorem}

\begin{example}
Consider the symmetric group $S_{3}$ as the Coxeter group $A_{2}$. Then one
can observe that
\begin{align*}
&  \mathbf{L}_{a}\mathbf{L}_{b}\\
&  =(1+2ab+2a^{2}b^{2}+a^{3}b^{3})\operatorname*{id}+(a+b+ab^{2}+a^{2}%
b+a^{3}b^{2}+a^{2}b^{3})(s_{1}+s_{2})\\
&  \ \ \ \ +(ab+a^{2}+b^{2}+a^{2}b^{2}+ab^{3}+a^{3}b)(s_{1}s_{2}+s_{2}%
s_{1})+(2ab^{2}+2a^{2}b+a^{3}+b^{3})w_{0}%
\end{align*}
with the notations of Example \ref{exa.LqSn}. This is visibly symmetric in $a$
and $b$, so it confirms Theorem \ref{thm.comm.dihe}.
\end{example}

We will prove Theorem \ref{thm.comm.dihe}, and many other results we shall
soon state, in the next section (Section \ref{sec.pfs}).

Next, even though the products $\mathbf{L}_{a}\mathbf{L}_{b}$ and
$\mathbf{L}_{b}\mathbf{L}_{a}$ are usually distinct for general $W$, we shall
show that certain coefficients of these products are equal. For this, we
introduce a notation:

\begin{definition}
\label{def.coeff}Let $W$ be a group. Let $a\in\mathbf{k}\left[  W\right]  $ be
an element of its group algebra. Let $w\in W$. Then, the $w$-coefficient of
$a$ will be denoted by $\left[  w\right]  a$.
\end{definition}

\begin{definition}
\label{def.Wls}Let $\left(  W,S\right)  $ be a Coxeter system with finite
Coxeter group $W$. Let $w\in W$. We say that the element $w$ is $W$%
\emph{-length-symmetric} if every commutative ring $\mathbf{k}$ and every
$a,b\in\mathbf{k}$ satisfy%
\begin{equation}
\left[  w\right]  \left(  \mathbf{L}_{a}\mathbf{L}_{b}\right)  =\left[
w\right]  \left(  \mathbf{L}_{b}\mathbf{L}_{a}\right)  . \label{eq.def.Wls.eq}%
\end{equation}

\end{definition}

Alternatively, $W$-length-symmetry can be characterized combinatorially as follows:

\begin{proposition}
\label{prop.Wls.comb}Let $\left(  W,S\right)  $ be a Coxeter system with
finite Coxeter group $W$. Let $w\in W$. Then, $w$ is $W$-length-symmetric if
and only if for every $p,q\in\mathbb{N}$, we have%
\begin{align*}
&  \left(  \text{number of }u\in W\text{ satisfying }\ell\left(  u\right)
=p\text{ and }\ell\left(  u^{-1}w\right)  =q\right) \\
&  =\left(  \text{number of }u\in W\text{ satisfying }\ell\left(  u\right)
=p\text{ and }\ell\left(  wu^{-1}\right)  =q\right)  .
\end{align*}

\end{proposition}

\begin{proof}
By Definition \ref{def.Lq}, we have $\mathbf{L}_{a}=\sum_{w\in W}%
a^{\ell\left(  w\right)  }w$ and $\mathbf{L}_{b}=\sum_{w\in W}b^{\ell\left(
w\right)  }w$ for all $a,b\in\mathbf{k}$. Thus, for all $a,b\in\mathbf{k}$, we
have
\begin{align*}
\mathbf{L}_{a}\mathbf{L}_{b}  &  =\left(  \sum_{w\in W}a^{\ell\left(
w\right)  }w\right)  \left(  \sum_{w\in W}b^{\ell\left(  w\right)  }w\right)
=\sum_{\left(  u,v\right)  \in W\times W}a^{\ell\left(  u\right)  }%
b^{\ell\left(  v\right)  }uv\ \ \ \ \ \ \ \ \ \ \text{and}\\
\mathbf{L}_{b}\mathbf{L}_{a}  &  =\left(  \sum_{w\in W}b^{\ell\left(
w\right)  }w\right)  \left(  \sum_{w\in W}a^{\ell\left(  w\right)  }w\right)
=\sum_{\left(  v,u\right)  \in W\times W}b^{\ell\left(  v\right)  }%
a^{\ell\left(  u\right)  }vu\\
&  =\sum_{\left(  v,u\right)  \in W\times W}a^{\ell\left(  u\right)  }%
b^{\ell\left(  v\right)  }vu.
\end{align*}
Hence, each $w\in W$ and $a,b\in\mathbf{k}$ satisfy%
\[
\left[  w\right]  \left(  \mathbf{L}_{a}\mathbf{L}_{b}\right)  =\sum
_{\substack{\left(  u,v\right)  \in W\times W;\\uv=w}}a^{\ell\left(  u\right)
}b^{\ell\left(  v\right)  }=\sum_{u\in W}a^{\ell\left(  u\right)  }%
b^{\ell\left(  u^{-1}w\right)  }%
\]
(here, we have substituted $\left(  u,u^{-1}w\right)  $ for $\left(
u,v\right)  $) and%
\[
\left[  w\right]  \left(  \mathbf{L}_{b}\mathbf{L}_{a}\right)  =\sum
_{\substack{\left(  v,u\right)  \in W\times W;\\vu=w}}a^{\ell\left(  u\right)
}b^{\ell\left(  v\right)  }=\sum_{u\in W}a^{\ell\left(  u\right)  }%
b^{\ell\left(  wu^{-1}\right)  }%
\]
(here, we have substituted $\left(  wu^{-1},u\right)  $ for $\left(
v,u\right)  $). Hence, the equality (\ref{eq.def.Wls.eq}) can be rewritten as%
\begin{equation}
\sum_{u\in W}a^{\ell\left(  u\right)  }b^{\ell\left(  u^{-1}w\right)  }%
=\sum_{u\in W}a^{\ell\left(  u\right)  }b^{\ell\left(  wu^{-1}\right)  }.
\label{eq.def.Wls.eq2}%
\end{equation}
This equality is a polynomial identity in $a$ and $b$. Thus, it holds for all
commutative rings $\mathbf{k}$ and all $a,b\in\mathbf{k}$ if and only if it
holds as a polynomial identity, i.e., if and only if we have%
\begin{equation}
\sum_{u\in W}X^{\ell\left(  u\right)  }Y^{\ell\left(  u^{-1}w\right)  }%
=\sum_{u\in W}X^{\ell\left(  u\right)  }Y^{\ell\left(  wu^{-1}\right)  }
\label{eq.def.Wls.eq3}%
\end{equation}
in the polynomial ring $\mathbb{Z}\left[  X,Y\right]  $ (because
(\ref{eq.def.Wls.eq2}) can be derived from (\ref{eq.def.Wls.eq3}) by
evaluating both sides at $X=a$ and $Y=b$, whereas (\ref{eq.def.Wls.eq3}) is
the particular case of (\ref{eq.def.Wls.eq2}) for $\mathbf{k}=\mathbb{Z}%
\left[  X,Y\right]  $ and $a=X$ and $b=Y$). But the identity
(\ref{eq.def.Wls.eq3}), in turn, is equivalent to saying that for every
$p,q\in\mathbb{N}$, we have%
\begin{align}
&  \left(  \text{number of }u\in W\text{ satisfying }\ell\left(  u\right)
=p\text{ and }\ell\left(  u^{-1}w\right)  =q\right) \nonumber\\
&  =\left(  \text{number of }u\in W\text{ satisfying }\ell\left(  u\right)
=p\text{ and }\ell\left(  wu^{-1}\right)  =q\right)  \label{eq.def.Wls.eq4}%
\end{align}
(since the former number is the $X^{p}Y^{q}$-coefficient on the left-hand side
of (\ref{eq.def.Wls.eq3}), whereas the latter number is the $X^{p}Y^{q}%
$-coefficient on the right-hand side of (\ref{eq.def.Wls.eq3})).

Combining these equivalences, we conclude that the equality
(\ref{eq.def.Wls.eq}) holds for all commutative rings $\mathbf{k}$ and all
$a,b\in\mathbf{k}$ if and only if the equality (\ref{eq.def.Wls.eq4}) holds
for every $p,q\in\mathbb{N}$. But the former equality is exactly the
definition of \textquotedblleft$W$-length-symmetric\textquotedblright; thus,
we have shown that $w$ is $W$-length-symmetric if and only if the equality
(\ref{eq.def.Wls.eq4}) holds for every $p,q\in\mathbb{N}$. This proves
Proposition \ref{prop.Wls.comb}.
\end{proof}

\begin{remark}
The characterization of $W$-length-symmetry in Proposition \ref{prop.Wls.comb}
makes perfect sense even if $W$ is infinite, as long as $S$ is finite (since
the number of $u\in W$ satisfying $\ell\left(  u\right)  =p$ is still finite
in this case). However, we will not concern ourselves with infinite Coxeter
groups here.
\end{remark}

\begin{remark}
\label{rmk.Wls.generic}As we have already noticed, the equality
(\ref{eq.def.Wls.eq2}) is a polynomial identity in $a$ and $b$. In other
words, the equality (\ref{eq.def.Wls.eq}) is a polynomial identity in $a$ and
$b$. Thus, in order for it to hold for all $\mathbf{k}$ and $a,b$, it suffices
that it hold when $\mathbf{k}=\mathbb{Q}$ and $a$ and $b$ are positive
integers (since two polynomials over $\mathbb{Q}$ are identical if they agree
on all positive integers). Hence, an element $w\in W$ is $W$-length-symmetric
if and only if it satisfies (\ref{eq.def.Wls.eq}) for $\mathbf{k}=\mathbb{Q}$
and for all positive integers $a$ and $b$. This gives yet another
combinatorial way to think of $W$-length-symmetry.
\end{remark}

\begin{example}
Consider the symmetric group $S_{n}$ as a Coxeter group. It is easy to see
(and follows from Corollary \ref{cor.Wls.dih} below) that each element of
$S_{n}$ is $S_{n}$-length-symmetric for all $n\leq3$. For $n=4$, all elements
of $S_{4}$ are $S_{4}$-length-symmetric except for the two permutations with
one-line notations $\left(  2,4,1,3\right)  $ and $\left(  3,1,4,2\right)  $.
These two permutations are precisely the two patterns that the \emph{separable
permutations} are defined to avoid. And we shall indeed see that every
separable permutation $w\in S_{n}$ is $S_{n}$-length-symmetric, although the
converse is not true. (We shall also see that every involution $w\in S_{n}$ is
$S_{n}$-length-symmetric, but not every involution is separable.)
\end{example}

\begin{remark}
\label{rmk.countSn}One might wonder how many $W$-length-symmetric elements
exist for a Coxeter group $W$. For instance, for each $n\in\mathbb{N}$, let
$\omega_{n}$ be the number of $S_{n}$-length-symmetric elements of the Coxeter
group $S_{n}$. Here are the first few values:%
\[%
\begin{tabular}
[c]{|c||c|c|c|c|c|c|c|c|}\hline
$n$ & $0$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$\\\hline
$\omega_{n}$ & $1$ & $1$ & $2$ & $6$ & $22$ & $94$ & $424$ & $2032$\\\hline
\end{tabular}
\ \
\]
This sequence is not in the OEIS as of 2026-06-05, even without $\omega_{0}$.
(Unverified code generated by GPT-5.5 additionally reports that $\omega
_{8}=9980$.)

Somewhat related is \href{https://oeis.org/A180389}{OEIS sequence A180389},
which counts the permutations $w\in S_{n}$ satisfying $\operatorname*{des}%
w=\operatorname*{des}\left(  w^{-1}\right)  $, where $\operatorname*{des}u$
denotes the number of descents of $u$ (that is, of simple reflections $s\in S$
satisfying $\ell\left(  us\right)  <\ell\left(  u\right)  $). Indeed, any
$W$-length-symmetric element $w\in S_{n}$ must satisfy $\operatorname*{des}%
w=\operatorname*{des}\left(  w^{-1}\right)  $, as can easily be seen from the
$p=1$ part of Proposition \ref{prop.Wls.comb}; but the converse is quite far
from truth.
\end{remark}

We shall now give several criteria for when elements of a Coxeter group are
$W$-length-symmetric. The first one is an easy consequence of Theorem
\ref{thm.comm.dihe}:

\begin{corollary}
\label{cor.Wls.dih}Let $\left(  W,S\right)  $ be a Coxeter system, where $W$
is finite and $\left\vert S\right\vert \leq2$. Then, each $w\in W$ is $W$-length-symmetric.
\end{corollary}

We shall now find $W$-length-symmetric elements in arbitrary finite Coxeter
groups. Some are fairly simple:

\begin{proposition}
\label{prop.Wls.inv}Let $\left(  W,S\right)  $ be a Coxeter system, where $W$
is finite. Then, each involution $w\in W$ (that is, each $w\in W$ satisfying
$w^{2}=1$) is $W$-length-symmetric.
\end{proposition}

\begin{proposition}
\label{prop.Wls.len2}Let $\left(  W,S\right)  $ be a Coxeter system, where $W$
is finite. Let $w\in W$ satisfy $\ell\left(  w\right)  \leq2$. Then, $w$ is
$W$-length-symmetric.
\end{proposition}

\begin{proposition}
\label{prop.Wls.inverse}Let $\left(  W,S\right)  $ be a Coxeter system, where
$W$ is finite. Let $w\in W$. Then, $w$ is $W$-length-symmetric if and only if
$w^{-1}$ is $W$-length-symmetric.
\end{proposition}

\begin{proposition}
\label{prop.Wls.ref}Let $\left(  W,S\right)  $ be a Coxeter system, where $W$
is finite. Let $w\in W$. Then, $w$ is $W$-length-symmetric if and only if
$w_{0}w$ is $W$-length-symmetric.
\end{proposition}

We can generalize Proposition \ref{prop.Wls.inv} from involutions to
\textquotedblleft twisted involutions\textquotedblright:

\begin{proposition}
\label{prop.Wls.twinv}Let $\left(  W,S\right)  $ be a Coxeter system, where
$W$ is finite. Let $\phi:W\rightarrow W$ be an automorphism of Coxeter groups
(that is, $\phi\left(  S\right)  =S$). Let $w\in W$ be such that $w^{-1}%
=\phi\left(  w\right)  $. Then, $w$ is $W$-length-symmetric.
\end{proposition}

\begin{example}
Let $\left(  W,S\right)  $ be a Coxeter system, where $W$ is finite. Then,
conjugation by $w_{0}$ (that is, the map $W\rightarrow W,\ w\mapsto
w_{0}ww_{0}^{-1}$) is an automorphism of Coxeter groups. The elements $w\in W$
satisfying $w^{-1}=w_{0}ww_{0}^{-1}$ are sometimes called the \emph{twisted
involutions}. Proposition \ref{prop.Wls.twinv} shows that they are $W$-length-symmetric.
\end{example}

Next, we shall show that $W$-length-symmetry does not really depend on $W$, in
the sense that an element of a standard parabolic subgroup $W_{I}$ of $W$ is
$W$-length-symmetric if and only if it is $W_{I}$-length-symmetric. First we
recall the definition of standard parabolic subgroups:

\begin{definition}
Let $\left(  W,S\right)  $ be a Coxeter system. If $I$ is a subset of $S$,
then $W_{I}$ denotes the subgroup of $W$ generated by $I$. This is called a
\emph{standard parabolic subgroup} of $W$. It is known (see, e.g.,
\cite[Proposition 9.5]{Lusztig}) that $W_{I}$ is again a Coxeter group, part
of the Coxeter system $\left(  W_{I},I\right)  $.
\end{definition}

\begin{proposition}
\label{prop.Wls.WI}Let $\left(  W,S\right)  $ be a Coxeter system, where $W$
is finite. Let $I$ be a subset of $S$. Let $w\in W_{I}$. Then, $w$ is
$W$-length-symmetric if and only if $w$ is $W_{I}$-length-symmetric.
\end{proposition}

Next, we discuss $W$-length-symmetry for reducible Coxeter groups. A Coxeter
group is said to be \emph{reducible} if it has an orthogonal decomposition;
this is defined as follows:

\begin{definition}
Let $\left(  W,S\right)  $ be a Coxeter system. Let $m\left(  i,j\right)  $
denote the order of $ij$ in $W$ for all $i,j\in S$. An \emph{orthogonal
decomposition} of $\left(  W,S\right)  $ means a pair $\left(  I,J\right)  $
of two disjoint nonempty subsets $I$ and $J$ of $S$ such that $S=I\cup J$ and
such that all $i\in I$ and $j\in J$ satisfy $m\left(  i,j\right)  =2$ (thus
$ij=ji$). We shall denote this decomposition as $S=I\oplus J$.
\end{definition}

It is easy to see that if $S=I\oplus J$ is an orthogonal decomposition, then
$W=W_{I}W_{J}\cong W_{I}\times W_{J}$.

\begin{proposition}
\label{prop.Wls.I+J}Let $\left(  W,S\right)  $ be a Coxeter system, where $W$
is finite. Let $S=I\oplus J$ be an orthogonal decomposition. Let $u\in W_{I}$
and $v\in W_{J}$. Assume that $u$ is $W_{I}$-length-symmetric and $v$ is
$W_{J}$-length-symmetric. Then, $uv$ is $W$-length-symmetric.
\end{proposition}

\begin{remark}
The converse of Proposition \ref{prop.Wls.I+J} is false. A large source of
counterexamples comes from Proposition \ref{prop.Wls.twinv}. Namely, let
$\left(  W,S\right)  $ be a Coxeter system that is a \textquotedblleft direct
product\textquotedblright\ of two isomorphic Coxeter systems -- i.e., that has
an orthogonal decomposition $S=I\oplus J$ and an isomorphism $\rho:\left(
W_{I},I\right)  \rightarrow\left(  W_{J},J\right)  $ of Coxeter groups. Let
$\phi:\left(  W,S\right)  \rightarrow\left(  W,S\right)  $ be the automorphism
of Coxeter groups that is induced by $\rho$ (that is, sends $I$ to $J$ via
$\rho$, and sends $J$ to $I$ via $\rho^{-1}$). Fix any $u\in W_{I}$, and let
$v:=\rho\left(  u^{-1}\right)  \in W_{J}$. Then, the element $uv\in W$
satisfies $\left(  uv\right)  ^{-1}=\phi\left(  uv\right)  $ (since the
definition of $\phi$ yields $\phi\left(  uv\right)  =\underbrace{\rho\left(
u\right)  }_{=v^{-1}}\underbrace{\rho^{-1}\left(  v\right)  }_{=u^{-1}}%
=v^{-1}u^{-1}=\left(  uv\right)  ^{-1}$), and thus is $W$-length-symmetric by
Proposition \ref{prop.Wls.twinv}. But $u$ is not necessarily $W_{I}%
$-length-symmetric, since it can be any element of $W_{I}$.
\end{remark}

\begin{remark}
\label{rmk.hecke}One can also define a variant of $W$-length-symmetry in the
Hecke algebra. Let $\left(  W,S\right)  $ be a Coxeter system with $W$ finite,
and let $\mathcal{H}$ be its Hecke algebra (see, e.g., \cite[\S 7.4]%
{Humphreys}) over the commutative ring $\mathbf{k}$ with a parameter
$q\in\mathbf{k}$. For any $r\in\mathbf{k}$, define
\[
\mathbf{L}_{r}^{\mathcal{H}}:=\sum_{w\in W}r^{\ell\left(  w\right)  }T_{w}%
\in\mathcal{H}.
\]
Then, an element $w\in W$ can be called \emph{Hecke-}$W$%
\emph{-length-symmetric} if the coefficient of $T_{w}$ in $\mathbf{L}%
_{a}^{\mathcal{H}}\mathbf{L}_{b}^{\mathcal{H}}$ equals the coefficient of
$T_{w}$ in $\mathbf{L}_{b}^{\mathcal{H}}\mathbf{L}_{a}^{\mathcal{H}}$ for
arbitrary $\mathbf{k}$ and $q,a,b\in\mathbf{k}$. This is a stronger property
than $W$-length-symmetry; in particular, the permutations $\left[
2431\right]  ,\ \left[  3241\right]  ,\ \left[  4132\right]  ,\ \left[
4213\right]  $ in $S_{4}$ (written in one-line notation) are $S_{4}%
$-length-symmetric but not Hecke-$S_{4}$-length-symmetric.

Finding Hecke analogues of the above results on $W$-length-symmetry appears
doable but not always trivial. An easy first step is generalizing Theorem
\ref{thm.comm.dihe}: If $\left\vert S\right\vert \leq2$, then%
\begin{equation}
\mathbf{L}_{a}^{\mathcal{H}}\mathbf{L}_{b}^{\mathcal{H}}=\mathbf{L}%
_{b}^{\mathcal{H}}\mathbf{L}_{a}^{\mathcal{H}}\ \ \ \ \ \ \ \ \ \ \text{for
all }q,a,b\in\mathbf{k}. \label{eq.thm.comm.dihe.hecke}%
\end{equation}
This can be proved in roughly the same way as we prove Theorem
\ref{thm.comm.dihe} below (viz., by replacing each $w$ by $T_{w}$ in Lemma
\ref{lem.dih.sti}, and slightly modifying its proof). Furthermore, Proposition
\ref{prop.Wls.inv}, Proposition \ref{prop.Wls.inverse} and Proposition
\ref{prop.Wls.twinv} also remain true if we replace \textquotedblleft%
$W$-length-symmetric\textquotedblright\ by \textquotedblleft Hecke-$W$%
-length-symmetric\textquotedblright, and the proof we give below generalizes
to this as well (since $\mathcal{H}$ also has an \textquotedblleft
antipode\textquotedblright, i.e., an involutive $\mathbf{k}$-algebra
anti-automorphism that sends each $T_{w}$ to $T_{w^{-1}}$). On the other hand,
a version of Proposition \ref{prop.Wls.ref} appears to be missing for
$\mathcal{H}$. The generalizability of Proposition \ref{prop.Wls.WI} is an
interesting question.
\end{remark}

\section{\label{sec.pfs}Proofs}

\subsection{Coefficients}

Before we step to the proofs of the above claims, we establish some basic facts.

We begin with some basic properties of the notation from Definition
\ref{def.coeff}:

\begin{lemma}
\label{lem.coeff}Let $W$ be a group. Let $w,g,h\in W$ and $a\in\mathbf{k}%
\left[  W\right]  $. Then,%
\[
\left[  w\right]  \left(  gah\right)  =\left[  g^{-1}wh^{-1}\right]  \left(
a\right)  .
\]

\end{lemma}

\begin{proof}
Both sides of the claimed equality are $\mathbf{k}$-linear in $a$. Thus, we
WLOG assume that $a$ is an element of the basis $W$ of the $\mathbf{k}$-module
$\mathbf{k}\left[  W\right]  $ (since $a$ is clearly a $\mathbf{k}$-linear
combination of these basis elements). In other words, $a\in W$. Hence, $gah\in
W$ as well, so that
\begin{equation}
\left[  w\right]  \left(  gah\right)  =%
\begin{cases}
1, & \text{if }w=gah;\\
0, & \text{otherwise}%
\end{cases}
\label{pf.lem.coeff.1}%
\end{equation}
(because any $p,q\in W$ satisfy $\left[  p\right]  q=%
\begin{cases}
1, & \text{if }p=q;\\
0, & \text{otherwise}%
\end{cases}
$) and similarly%
\begin{equation}
\left[  g^{-1}wh^{-1}\right]  \left(  a\right)  =%
\begin{cases}
1, & \text{if }g^{-1}wh^{-1}=a;\\
0, & \text{otherwise.}%
\end{cases}
\label{pf.lem.coeff.2}%
\end{equation}
However, $w=gah$ holds if and only if $g^{-1}wh^{-1}=a$. Hence, the right-hand
sides of the equalities (\ref{pf.lem.coeff.1}) and (\ref{pf.lem.coeff.2}) are
equal. Thus, their left-hand sides are equal as well. In other words, $\left[
w\right]  \left(  gah\right)  =\left[  g^{-1}wh^{-1}\right]  \left(  a\right)
$. This proves Lemma \ref{lem.coeff}.
\end{proof}

\subsection{Dihedral groups}

Dihedral groups (Coxeter groups with two generators) are simple but form a
crucial building block of Coxeter group theory. The following lemma is likely
known in some form, but we could not find it in the literature.

\begin{lemma}
\label{lem.dih.sti}Let $\left(  W,S\right)  $ be a Coxeter system, where
$S=\left\{  s,t\right\}  $ is a two-element set. (Thus, $W$ is a dihedral
group.) For each $i\in\mathbb{N}$, we let $\left(  st\right)  _{i}$ denote the
product $ststs\cdots$ with $i$ factors (that is, $\left(  st\right)  ^{i/2}$
if $i$ is even, and $\left(  st\right)  ^{\left(  i-1\right)  /2}s$ if $i$ is
odd), and we let $\left(  ts\right)  _{i}$ denote the product $tstst\cdots$
with $i$ factors. Let $B$ be the $\mathbf{k}$-subalgebra of the group algebra
$\mathbf{k}\left[  W\right]  $ generated by $s+t$. Then, each $i\in\mathbb{N}$
satisfies%
\[
\left(  st\right)  _{i}+\left(  ts\right)  _{i}\in B.
\]

\end{lemma}

\begin{proof}
We prove this by strong induction on $i$. The \textit{base cases} $i=0$ and
$i=1$ are obvious (since $\left(  st\right)  _{0}+\left(  ts\right)
_{0}=1+1=2$ and $\left(  st\right)  _{1}+\left(  ts\right)  _{1}=s+t$). For
the \textit{induction step}, we observe that each $i\geq2$ satisfies%
\begin{align*}
&  \left(  s+t\right)  \left(  \left(  st\right)  _{i-1}+\left(  ts\right)
_{i-1}\right) \\
&  =s\underbrace{\left(  st\right)  _{i-1}}_{=s\left(  ts\right)  _{i-2}%
}+\underbrace{s\left(  ts\right)  _{i-1}}_{=\left(  st\right)  _{i}%
}+\underbrace{t\left(  st\right)  _{i-1}}_{=\left(  ts\right)  _{i}%
}+t\underbrace{\left(  ts\right)  _{i-1}}_{=t\left(  st\right)  _{i-2}}\\
&  =\underbrace{ss}_{=s^{2}=1}\left(  ts\right)  _{i-2}+\left(  st\right)
_{i}+\left(  ts\right)  _{i}+\underbrace{tt}_{=t^{2}=1}\left(  st\right)
_{i-2}\\
&  =\left(  ts\right)  _{i-2}+\left(  st\right)  _{i}+\left(  ts\right)
_{i}+\left(  st\right)  _{i-2}=\left(  st\right)  _{i}+\left(  ts\right)
_{i}+\left(  st\right)  _{i-2}+\left(  ts\right)  _{i-2}%
\end{align*}
and thus%
\begin{align*}
\left(  st\right)  _{i}+\left(  ts\right)  _{i}  &  =\underbrace{\left(
s+t\right)  }_{\in B}\underbrace{\left(  \left(  st\right)  _{i-1}+\left(
ts\right)  _{i-1}\right)  }_{\substack{\in B\\\text{(by the induction
hypothesis)}}}-\underbrace{\left(  \left(  st\right)  _{i-2}+\left(
ts\right)  _{i-2}\right)  }_{\substack{\in B\\\text{(by the induction
hypothesis)}}}\\
&  \in B\ \ \ \ \ \ \ \ \ \ \left(  \text{since }B\text{ is a subalgebra}%
\right)  .
\end{align*}
This completes the induction step, and thus the proof of the lemma.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.comm.dihe}.]Let $a,b\in\mathbf{k}$. We must prove
that $\mathbf{L}_{a}\mathbf{L}_{b}=\mathbf{L}_{b}\mathbf{L}_{a}$.

If $\left\vert S\right\vert \leq1$, then the $\mathbf{k}$-algebra
$\mathbf{k}\left[  W\right]  $ is commutative (since it is generated by $S$),
and thus the claim $\mathbf{L}_{a}\mathbf{L}_{b}=\mathbf{L}_{b}\mathbf{L}_{a}$
holds for obvious reasons. Thus, we WLOG assume that $\left\vert S\right\vert
>1$ from now on.

Hence, $\left\vert S\right\vert =2$ (since we assumed that $\left\vert
S\right\vert \leq2$). Write this $2$-element set $S$ as $S=\left\{
s,t\right\}  $.

Use the notations from Lemma \ref{lem.dih.sti}. In particular, define the
$\mathbf{k}$-subalgebra $B$ of $\mathbf{k}\left[  W\right]  $ as in Lemma
\ref{lem.dih.sti}. Note that this algebra $B$ is commutative, since it is
generated by a single element.

However, $W$ is a dihedral group, so it has a well-known description: If we
let $m$ denote the order of $st$ in the group $W$, then the elements of $W$
are precisely the products $\left(  st\right)  _{i}$ and $\left(  ts\right)
_{i}$ for all $i\in\left\{  0,1,\ldots,m\right\}  $, which are furthermore all
distinct except that $\left(  st\right)  _{0}=\left(  ts\right)  _{0}$ and
$\left(  st\right)  _{m}=\left(  ts\right)  _{m}$. Moreover, the products
$\left(  st\right)  _{i}$ and $\left(  ts\right)  _{i}$ have length $i$ for
all $i\in\left\{  0,1,\ldots,m\right\}  $. Altogether, we see that the
elements of $W$ (listed without repetitions) are%
\[
\underbrace{1}_{\text{length }0},\ \underbrace{\left(  st\right)
_{1},\ \left(  ts\right)  _{1}}_{\text{length }1},\ \underbrace{\left(
st\right)  _{2},\ \left(  ts\right)  _{2}}_{\text{length }2},\ \ldots
,\ \underbrace{\left(  st\right)  _{m-1},\ \left(  ts\right)  _{m-1}%
}_{\text{length }m-1},\ \underbrace{\left(  st\right)  _{m}}_{\text{length }%
m}.
\]
Thus, the definition $\mathbf{L}_{a}=\sum_{w\in W}a^{\ell\left(  w\right)  }w$
of $\mathbf{L}_{a}$ can be rewritten as%
\begin{align}
\mathbf{L}_{a}  &  =1+\sum_{i=1}^{m-1}\left(  a^{i}\left(  st\right)
_{i}+a^{i}\left(  ts\right)  _{i}\right)  +a^{m}\left(  st\right)
_{m}\nonumber\\
&  =1+\sum_{i=1}^{m-1}a^{i}\left(  \left(  st\right)  _{i}+\left(  ts\right)
_{i}\right)  +a^{m}\left(  st\right)  _{m}. \label{eq.thm.comm.dihe.1}%
\end{align}
Let us WLOG assume that $\mathbf{k}$ is a polynomial ring over $\mathbb{Q}$
(we can do this, since we are proving a polynomial identity, so we can assume
that $\mathbf{k}$, $a$ and $b$ are generic, as explained in Remark
\ref{rmk.Wls.generic}). Thus, $2$ is invertible in $\mathbf{k}$. Hence, in
(\ref{eq.thm.comm.dihe.1}), we can rewrite $\left(  st\right)  _{m}$ as
$\dfrac{\left(  st\right)  _{m}+\left(  ts\right)  _{m}}{2}$ (since $\left(
st\right)  _{m}=\left(  ts\right)  _{m}$), and thus (\ref{eq.thm.comm.dihe.1})
becomes%
\[
\mathbf{L}_{a}=1+\sum_{i=1}^{m-1}a^{i}\left(  \left(  st\right)  _{i}+\left(
ts\right)  _{i}\right)  +a^{m}\dfrac{\left(  st\right)  _{m}+\left(
ts\right)  _{m}}{2}.
\]
This shows that $\mathbf{L}_{a}$ belongs to $B$ (since Lemma \ref{lem.dih.sti}
shows that all the sums $\left(  st\right)  _{i}+\left(  ts\right)  _{i}$
belong to $B$). Similarly, $\mathbf{L}_{b}$ belongs to $B$ as well. Hence,
$\mathbf{L}_{a}$ and $\mathbf{L}_{b}$ commute (since $B$ is commutative). That
is, $\mathbf{L}_{a}\mathbf{L}_{b}=\mathbf{L}_{b}\mathbf{L}_{a}$. Theorem
\ref{thm.comm.dihe} is thus proved.
\end{proof}

\begin{proof}
[Proof of Corollary \ref{cor.Wls.dih}.]Let $w\in W$. For all $a,b\in
\mathbf{k}$, we have $\mathbf{L}_{a}\mathbf{L}_{b}=\mathbf{L}_{b}%
\mathbf{L}_{a}$ by Theorem \ref{thm.comm.dihe}, and therefore $\left[
w\right]  \left(  \mathbf{L}_{a}\mathbf{L}_{b}\right)  =\left[  w\right]
\left(  \mathbf{L}_{b}\mathbf{L}_{a}\right)  $. In other words, $w$ is
$W$-length-symmetric. This proves Corollary \ref{cor.Wls.dih}.
\end{proof}

\subsection{The antipode, inverses and involutions}

For any group $G$, there is a $\mathbf{k}$-linear map $\mathbf{k}\left[
G\right]  \rightarrow\mathbf{k}\left[  G\right]  $ that sends each group
element $g\in G$ to $g^{-1}$. This map is called the \emph{antipode} of the
group algebra $\mathbf{k}\left[  G\right]  $, and will be denoted by $a\mapsto
a^{\ast}$. It is an involution (i.e., each $a\in\mathbf{k}\left[  G\right]  $
satisfies $a^{\ast\ast}=a$), and is a $\mathbf{k}$-algebra anti-morphism
(i.e., each $a,b\in\mathbf{k}\left[  G\right]  $ satisfy $\left(  ab\right)
^{\ast}=b^{\ast}a^{\ast}$). Thus, it is a $\mathbf{k}$-algebra
anti-isomorphism. Clearly, all $w\in G$ and $a\in\mathbf{k}\left[  G\right]  $
satisfy%
\begin{equation}
\left[  w\right]  \left(  a\right)  =\left[  w^{-1}\right]  \left(  a^{\ast
}\right)  . \label{eq.antipode.coeff}%
\end{equation}


\begin{lemma}
\label{lem.W-S}Let $\left(  W,S\right)  $ be a Coxeter system, where $W$ is
finite. Let $q\in\mathbf{k}$. Then, $\mathbf{L}_{q}^{\ast}=\mathbf{L}_{q}$.
\end{lemma}

\begin{proof}
This is clear, since each $w\in W$ satisfies $\ell\left(  w^{-1}\right)
=\ell\left(  w\right)  $.
\end{proof}

\begin{lemma}
\label{lem.winv}Let $\left(  W,S\right)  $ be a Coxeter system, where $W$ is
finite. Let $a,b\in\mathbf{k}$ and $w\in W$. Then, $\left[  w\right]  \left(
\mathbf{L}_{a}\mathbf{L}_{b}\right)  =\left[  w^{-1}\right]  \left(
\mathbf{L}_{b}\mathbf{L}_{a}\right)  $.
\end{lemma}

\begin{proof}
We have $\left(  \mathbf{L}_{a}\mathbf{L}_{b}\right)  ^{\ast}=\mathbf{L}%
_{b}^{\ast}\mathbf{L}_{a}^{\ast}$ (since the antipode is a $\mathbf{k}%
$-algebra anti-morphism). But Lemma \ref{lem.W-S} yields $\mathbf{L}_{a}%
^{\ast}=\mathbf{L}_{a}$ and $\mathbf{L}_{b}^{\ast}=\mathbf{L}_{b}$. Hence,
$\left(  \mathbf{L}_{a}\mathbf{L}_{b}\right)  ^{\ast}=\underbrace{\mathbf{L}%
_{b}^{\ast}}_{=\mathbf{L}_{b}}\underbrace{\mathbf{L}_{a}^{\ast}}%
_{=\mathbf{L}_{a}}=\mathbf{L}_{b}\mathbf{L}_{a}$. Now,
(\ref{eq.antipode.coeff}) yields%
\[
\left[  w\right]  \left(  \mathbf{L}_{a}\mathbf{L}_{b}\right)  =\left[
w^{-1}\right]  \left(  \left(  \mathbf{L}_{a}\mathbf{L}_{b}\right)  ^{\ast
}\right)  =\left[  w^{-1}\right]  \left(  \mathbf{L}_{b}\mathbf{L}_{a}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \mathbf{L}_{a}\mathbf{L}%
_{b}\right)  ^{\ast}=\mathbf{L}_{b}\mathbf{L}_{a}\right)  ,
\]
and thus Lemma \ref{lem.winv} is proved.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.Wls.inverse}.]Let $a,b\in\mathbf{k}$ be
arbitrary. Then,%
\begin{align*}
\left[  w\right]  \left(  \mathbf{L}_{a}\mathbf{L}_{b}\right)   &  =\left[
w^{-1}\right]  \left(  \mathbf{L}_{b}\mathbf{L}_{a}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma \ref{lem.winv}}\right)
,\ \ \ \ \ \ \ \ \ \ \text{and}\\
\left[  w\right]  \left(  \mathbf{L}_{b}\mathbf{L}_{a}\right)   &  =\left[
w^{-1}\right]  \left(  \mathbf{L}_{a}\mathbf{L}_{b}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma \ref{lem.winv}, with the roles of
}a\text{ and }b\text{ swapped}\right)  .
\end{align*}
Hence, the equality $\left[  w\right]  \left(  \mathbf{L}_{a}\mathbf{L}%
_{b}\right)  =\left[  w\right]  \left(  \mathbf{L}_{b}\mathbf{L}_{a}\right)  $
holds if and only if the equality \newline$\left[  w^{-1}\right]  \left(
\mathbf{L}_{b}\mathbf{L}_{a}\right)  =\left[  w^{-1}\right]  \left(
\mathbf{L}_{a}\mathbf{L}_{b}\right)  $ holds. But the former equality (stated
for all $\mathbf{k}$ and $a,b\in\mathbf{k}$) says that $w$ is $W$%
-length-symmetric, whereas the latter equality says that $w^{-1}$ is
$W$-length-symmetric. Hence, $w$ is $W$-length-symmetric if and only if
$w^{-1}$ is $W$-length-symmetric. This proves Proposition
\ref{prop.Wls.inverse}.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.Wls.inv}.]Let $w\in W$ be an involution, and
$a,b\in\mathbf{k}$ be arbitrary. Then, $w$ is an involution, so that
$w^{-1}=w$. However, Lemma \ref{lem.winv} yields%
\[
\left[  w\right]  \left(  \mathbf{L}_{a}\mathbf{L}_{b}\right)  =\left[
w^{-1}\right]  \left(  \mathbf{L}_{b}\mathbf{L}_{a}\right)  =\left[  w\right]
\left(  \mathbf{L}_{b}\mathbf{L}_{a}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }w^{-1}=w\right)  .
\]
This shows that $w$ is $W$-length-symmetric. This proves Proposition
\ref{prop.Wls.inv}.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.Wls.twinv}.]Forget that we fixed $w$. The
group automorphism $\phi$ of $W$ induces (by linearity) a $\mathbf{k}$-algebra
automorphism $\mathbf{k}\left[  \phi\right]  $ of $\mathbf{k}\left[  W\right]
$, which sends each $w\in W$ to $\phi\left(  w\right)  $.

Moreover, the automorphism $\phi$ of $W$ is a Coxeter group automorphism.
Thus, it preserves the length of any element $w\in W$; that is, we have%
\begin{equation}
\ell\left(  \phi\left(  w\right)  \right)  =\ell\left(  w\right)
\ \ \ \ \ \ \ \ \ \ \text{for each }w\in W. \label{pf.prop.WIs.twinv.pres}%
\end{equation}
Thus, for each $q\in\mathbf{k}$, we have%
\begin{align*}
\left(  \mathbf{k}\left[  \phi\right]  \right)  \left(  \mathbf{L}_{q}\right)
&  =\left(  \mathbf{k}\left[  \phi\right]  \right)  \left(  \sum_{w\in
W}q^{\ell\left(  w\right)  }w\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\mathbf{L}_{q}=\sum_{w\in W}q^{\ell\left(  w\right)  }w\right) \\
&  =\sum_{w\in W}\underbrace{q^{\ell\left(  w\right)  }}_{\substack{=q^{\ell
\left(  \phi\left(  w\right)  \right)  }\\\text{(since
(\ref{pf.prop.WIs.twinv.pres}) yields }\ell\left(  w\right)  =\ell\left(
\phi\left(  w\right)  \right)  }}\phi\left(  w\right)  =\sum_{w\in W}%
q^{\ell\left(  \phi\left(  w\right)  \right)  }\phi\left(  w\right) \\
&  =\sum_{w\in W}q^{\ell\left(  w\right)  }w\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }w\text{ for }\phi\left(  w\right)  \text{
in}\\
\text{the sum, since }\phi:W\rightarrow W\text{ is a bijection}%
\end{array}
\right) \\
&  =\mathbf{L}_{q}.
\end{align*}
In other words, $\mathbf{k}\left[  \phi\right]  $ fixes $\mathbf{L}_{q}$ for
each $q\in\mathbf{k}$.

Let $a,b\in\mathbf{k}$ be arbitrary. Thus, the $\mathbf{k}$-algebra
automorphism $\mathbf{k}\left[  \phi\right]  $ of $\mathbf{k}\left[  W\right]
$ fixes the elements $\mathbf{L}_{a}$ and $\mathbf{L}_{b}$ (since it fixes
$\mathbf{L}_{q}$ for each $q\in\mathbf{k}$). Being a $\mathbf{k}$-algebra
automorphism, it therefore also fixes the product $\mathbf{L}_{b}%
\mathbf{L}_{a}$. In other words,%
\begin{equation}
\left(  \mathbf{k}\left[  \phi\right]  \right)  \left(  \mathbf{L}%
_{b}\mathbf{L}_{a}\right)  =\mathbf{L}_{b}\mathbf{L}_{a}.
\label{pf.prop.Wls.twinv.0}%
\end{equation}
Moreover, by its very definition, it satisfies%
\begin{equation}
\left[  \phi\left(  w\right)  \right]  \left(  \left(  \mathbf{k}\left[
\phi\right]  \right)  \left(  \alpha\right)  \right)  =\left[  w\right]
\left(  \alpha\right)  \label{pf.prop.Wls.twinv.1}%
\end{equation}
for all $w\in W$ and $\alpha\in\mathbf{k}\left[  W\right]  $.

Now, let $w\in W$ be such that $w^{-1}=\phi\left(  w\right)  $. Now, Lemma
\ref{lem.winv} yields%
\begin{align*}
\left[  w\right]  \left(  \mathbf{L}_{a}\mathbf{L}_{b}\right)   &  =\left[
w^{-1}\right]  \left(  \mathbf{L}_{b}\mathbf{L}_{a}\right) \\
&  =\left[  \phi\left(  w\right)  \right]  \underbrace{\left(  \mathbf{L}%
_{b}\mathbf{L}_{a}\right)  }_{\substack{=\left(  \mathbf{k}\left[
\phi\right]  \right)  \left(  \mathbf{L}_{b}\mathbf{L}_{a}\right)  \\\text{(by
(\ref{pf.prop.Wls.twinv.0}))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
w^{-1}=\phi\left(  w\right)  \right) \\
&  =\left[  \phi\left(  w\right)  \right]  \left(  \left(  \mathbf{k}\left[
\phi\right]  \right)  \left(  \mathbf{L}_{b}\mathbf{L}_{a}\right)  \right)
=\left[  w\right]  \left(  \mathbf{L}_{b}\mathbf{L}_{a}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.Wls.twinv.1})}\right)  .
\end{align*}
This shows that $w$ is $W$-length-symmetric. This proves Proposition
\ref{prop.Wls.twinv}.
\end{proof}

\subsection{Cosets and parabolic subgroups}

Our next lemma is concerned with cosets and their representatives. Due to the
perennial vexatiousness of their chirality, we recall their definitions:

\begin{definition}
Let $H$ be a subgroup of a group $G$.

\begin{itemize}
\item The \emph{left cosets} of $H$ in $G$ are the subsets of the form $uH$
for $u\in G$. The set of these left cosets is denoted by $G/H$.

A \emph{left transversal} of $H$ in $G$ means a subset $L$ of $G$ that
intersects each of these left cosets in exactly one element (i.e., satisfies
$\left\vert uH\cap L\right\vert =1$ for each $u\in G$). Equivalently, it is a
subset $L$ of $G$ with the property that each $g\in G$ can be written uniquely
as $xh$ with $h\in H$ and $x\in L$.

\item The \emph{right cosets} of $H$ in $G$ are the subsets of the form $Hu$
for $u\in G$. The set of these right cosets is denoted by $H\backslash G$.

A \emph{right transversal} of $H$ in $G$ means a subset $R$ of $G$ that
intersects each of these right cosets in exactly one element (i.e., satisfies
$\left\vert Hu\cap R\right\vert =1$ for each $u\in G$). Equivalently, it is a
subset $R$ of $G$ with the property that each $g\in G$ can be written uniquely
as $hx$ with $h\in H$ and $x\in R$.
\end{itemize}
\end{definition}

The left and right versions here are intimately related: If $L$ is a left
transversal of $H$ in $G$, then $\left\{  x^{-1}\ \mid\ x\in L\right\}  $ is a
right transversal of $H$ in $G$. An analogous statement holds with
\textquotedblleft left\textquotedblright\ and \textquotedblleft
right\textquotedblright\ interchanged.

\begin{lemma}
\label{lem.coset}Let $G$ be a finite group, and let $H$ be a subgroup of $G$.
Let $R$ be a right transversal of $H$ in $G$. For each $x\in R$, let
$p_{x},q_{x}\in\mathbf{k}$ be two scalars. Set%
\[
p=\sum_{x\in R}p_{x}x\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ q=\sum
_{x\in R}q_{x}x^{-1}%
\]
in $\mathbf{k}\left[  G\right]  $. Let $w\in H$ and $\alpha,\beta\in
\mathbf{k}\left[  H\right]  $ (where $\mathbf{k}\left[  H\right]  $ is
embedded in $\mathbf{k}\left[  G\right]  $ in the obvious way). Then,%
\begin{equation}
\left[  w\right]  \left(  \alpha pq\beta\right)  =\left(  \sum_{x\in R}%
p_{x}q_{x}\right)  \left[  w\right]  \left(  \alpha\beta\right)  .
\label{eq.lem.coset.eq}%
\end{equation}

\end{lemma}

\begin{proof}
The claim is $\mathbf{k}$-linear in each of $\alpha$ and $\beta$. Thus, we
WLOG assume that $\alpha,\beta\in H$ (since $\alpha,\beta\in\mathbf{k}\left[
H\right]  $ are clearly $\mathbf{k}$-linear combinations of elements of $H$).
Furthermore, let us set $u:=\alpha^{-1}w\beta^{-1}\in H$ (since $w,\alpha
,\beta\in H$).

Hence, $w,\alpha,\beta\in H\subseteq G$. Thus, Lemma \ref{lem.coeff} (applied
to $\alpha$, $\beta$ and $pq$ instead of $g$, $h$ and $a$) yields%
\begin{equation}
\left[  w\right]  \left(  \alpha pq\beta\right)  =\left[  \alpha^{-1}%
w\beta^{-1}\right]  \left(  pq\right)  =\left[  u\right]  \left(  pq\right)
\label{pf.lem.coset.1}%
\end{equation}
(since $\alpha^{-1}w\beta^{-1}=u$). Moreover, Lemma \ref{lem.coeff} (applied
to $\alpha$, $\beta$ and $1$ instead of $g$, $h$ and $a$) yields%
\begin{equation}
\left[  w\right]  \left(  \alpha\beta\right)  =\left[  \alpha^{-1}w\beta
^{-1}\right]  1=\left[  u\right]  1 \label{pf.lem.coset.2}%
\end{equation}
(since $\alpha^{-1}w\beta^{-1}=u$). Using (\ref{pf.lem.coset.1}) and
(\ref{pf.lem.coset.2}), we can rewrite our claim (\ref{eq.lem.coset.eq}) as%
\begin{equation}
\left[  u\right]  \left(  pq\right)  =\left(  \sum_{x\in R}p_{x}q_{x}\right)
\left[  u\right]  \left(  1\right)  . \label{pf.lem.coset.goalrew}%
\end{equation}
So it remains to prove (\ref{pf.lem.coset.goalrew}).

But the definitions of $p$ and $q$ yield%
\begin{align*}
pq  &  =\left(  \sum_{x\in R}p_{x}x\right)  \left(  \sum_{x\in R}q_{x}%
x^{-1}\right)  =\sum_{\left(  x,y\right)  \in R\times R}p_{x}q_{y}xy^{-1}\\
&  =\sum_{\substack{\left(  x,y\right)  \in R\times R;\\x=y}}p_{x}q_{y}%
xy^{-1}+\sum_{\substack{\left(  x,y\right)  \in R\times R;\\x\neq y}%
}p_{x}q_{y}xy^{-1}%
\end{align*}
(here, we have split the sum into the part with $x=y$ and the part with $x\neq
y$). In view of%
\[
\sum_{\substack{\left(  x,y\right)  \in R\times R;\\x=y}}p_{x}q_{y}%
xy^{-1}=\sum_{x\in R}p_{x}q_{x}\underbrace{xx^{-1}}_{=1}=\sum_{x\in R}%
p_{x}q_{x}1,
\]
we can rewrite this as%
\begin{equation}
pq=\sum_{x\in R}p_{x}q_{x}1+\sum_{\substack{\left(  x,y\right)  \in R\times
R;\\x\neq y}}p_{x}q_{y}xy^{-1}. \label{pf.lem.coset.6}%
\end{equation}
Comparing coefficients of $u$ on both sides of this equality, we obtain%
\begin{equation}
\left[  u\right]  \left(  pq\right)  =\sum_{x\in R}p_{x}q_{x}\left[  u\right]
1+\sum_{\substack{\left(  x,y\right)  \in R\times R;\\x\neq y}}p_{x}%
q_{y}\left[  u\right]  \left(  xy^{-1}\right)  . \label{pf.lem.coset.7}%
\end{equation}


However, we claim that if two elements $x,y\in R$ satisfy $x\neq y$, then
\begin{equation}
\left[  u\right]  \left(  xy^{-1}\right)  =0. \label{pf.lem.coset.8}%
\end{equation}


[\textit{Proof:} Let $x,y\in R$ be two elements satisfying $x\neq y$. Then,
$x$ and $y$ are two distinct elements of $R$. Since $R$ is a right transversal
of $H$, this entails that the right cosets $Hx$ and $Hy$ are distinct. Thus,
the two elements $u$ and $xy^{-1}$ of $G$ are distinct (because if they were
equal, then we would have $xy^{-1}=u\in H$, so that $x\in Hy$ and therefore
$Hx=Hy$, contradicting the distinctness of $Hx$ and $Hy$). Therefore, $\left[
u\right]  \left(  xy^{-1}\right)  =0$ (because $\left[  g\right]  h=0$ for any
two distinct elements $g$ and $h$ of $G$). This proves (\ref{pf.lem.coset.8}).]

Now, (\ref{pf.lem.coset.7}) becomes%
\begin{align*}
\left[  u\right]  \left(  pq\right)   &  =\sum_{x\in R}p_{x}q_{x}\left[
u\right]  1+\sum_{\substack{\left(  x,y\right)  \in R\times R;\\x\neq y}%
}p_{x}q_{y}\underbrace{\left[  u\right]  \left(  xy^{-1}\right)
}_{\substack{=0\\\text{(by (\ref{pf.lem.coset.8}))}}}=\sum_{x\in R}p_{x}%
q_{x}\left[  u\right]  1\\
&  =\left(  \sum_{x\in R}p_{x}q_{x}\right)  \left[  u\right]  \left(
1\right)  .
\end{align*}
This proves (\ref{pf.lem.coset.goalrew}). Thus, Lemma \ref{lem.coset} is proved.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.Wls.WI}.]Given any $q\in\mathbf{k}$, let us
denote the element $\mathbf{L}_{q}\left(  W\right)  \in\mathbf{k}\left[
W\right]  $ by $\mathbf{L}_{q}$, while the analogous element $\mathbf{L}%
_{q}\left(  W_{I}\right)  \in\mathbf{k}\left[  W_{I}\right]  $ shall be
denoted by $\mathbf{L}_{q}^{\prime}$. Thus, $w$ is $W$-length-symmetric if and
only if all $\mathbf{k}$ and $a,b\in\mathbf{k}$ satisfy $\left[  w\right]
\left(  \mathbf{L}_{a}\mathbf{L}_{b}\right)  =\left[  w\right]  \left(
\mathbf{L}_{b}\mathbf{L}_{a}\right)  $, whereas $w$ is $W_{I}$%
-length-symmetric if and only if all $\mathbf{k}$ and $a,b\in\mathbf{k}$
satisfy $\left[  w\right]  \left(  \mathbf{L}_{a}^{\prime}\mathbf{L}%
_{b}^{\prime}\right)  =\left[  w\right]  \left(  \mathbf{L}_{b}^{\prime
}\mathbf{L}_{a}^{\prime}\right)  $. In both cases, we can restrict ourselves
to the \textquotedblleft generic\textquotedblright\ case, i.e., to the case
when $\mathbf{k}$ is the polynomial ring $\mathbb{Z}\left[  a,b\right]  $ and
when $a$ and $b$ are the two indeterminates (as in the proof of Proposition
\ref{prop.Wls.comb}). Let us do so. Thus, $w$ is $W$-length-symmetric if and
only if $\left[  w\right]  \left(  \mathbf{L}_{a}\mathbf{L}_{b}\right)
=\left[  w\right]  \left(  \mathbf{L}_{b}\mathbf{L}_{a}\right)  $, whereas $w$
is $W_{I}$-length-symmetric if and only if $\left[  w\right]  \left(
\mathbf{L}_{a}^{\prime}\mathbf{L}_{b}^{\prime}\right)  =\left[  w\right]
\left(  \mathbf{L}_{b}^{\prime}\mathbf{L}_{a}^{\prime}\right)  $.

Each right coset of $W_{I}$ in $W$ has a unique representative of minimum
length (by \cite[Lemma 9.7 (a)]{Lusztig}). Let $R$ be the set of all these
representatives. Thus, $R$ is a right transversal of $W_{I}$ in $W$. In other
words, each $g\in W$ can be written uniquely as $hx$ with $h\in W_{I}$ and
$x\in R$. That is, the map $W_{I}\times R\rightarrow W,\ \left(  h,x\right)
\mapsto hx$ is a bijection. Moreover, \cite[Lemma 9.7 (b)]{Lusztig} says that
any $x\in R$ and any $h\in W_{I}$ satisfy%
\begin{equation}
\ell\left(  hx\right)  =\ell\left(  h\right)  +\ell\left(  x\right)
\label{eq.prop.Wls.WI.l=l+l}%
\end{equation}
(since $x\in R$ means that $x$ is a representative of minimum length for some
right coset $W_{I}a$ of $W_{I}$ in $W$). Set%
\[
\mathbf{X}_{a}:=\sum_{x\in R}a^{\ell\left(  x\right)  }%
x\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \mathbf{X}_{b}:=\sum_{x\in
R}b^{\ell\left(  x\right)  }x.
\]
Then, the definition of $\mathbf{L}_{a}=\mathbf{L}_{a}\left(  W\right)  $
yields%
\begin{align}
\mathbf{L}_{a}  &  =\sum_{w\in W}a^{\ell\left(  w\right)  }w=\underbrace{\sum
_{\left(  h,x\right)  \in W_{I}\times R}}_{=\sum_{h\in W_{I}}\ \ \sum_{x\in
R}}\underbrace{a^{\ell\left(  hx\right)  }}_{\substack{=a^{\ell\left(
h\right)  +\ell\left(  x\right)  }\\\text{(by (\ref{eq.prop.Wls.WI.l=l+l}))}%
}}hx\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }hx\text{ for }w\text{ in the sum,}\\
\text{since the map }W_{I}\times R\rightarrow W,\ \left(  h,x\right)  \mapsto
hx\text{ is a bijection}%
\end{array}
\right) \nonumber\\
&  =\sum_{h\in W_{I}}\ \ \sum_{x\in R}a^{\ell\left(  h\right)  +\ell\left(
x\right)  }hx=\underbrace{\left(  \sum_{h\in W_{I}}a^{\ell\left(  h\right)
}h\right)  }_{\substack{=\mathbf{L}_{a}^{\prime}\\\text{(by the definition of
}\mathbf{L}_{a}^{\prime}=\mathbf{L}_{a}\left(  W_{I}\right)  \text{)}%
}}\ \ \underbrace{\left(  \sum_{x\in R}a^{\ell\left(  x\right)  }x\right)
}_{\substack{=\mathbf{X}_{a}\\\text{(by the definition of }\mathbf{X}%
_{a}\text{)}}}\nonumber\\
&  =\mathbf{L}_{a}^{\prime}\mathbf{X}_{a}. \label{pf.prop.Wls.WI.3}%
\end{align}
Similarly, $\mathbf{L}_{b}=\mathbf{L}_{b}^{\prime}\mathbf{X}_{b}$. Applying
the antipode to this latter equality, we obtain%
\begin{equation}
\mathbf{L}_{b}^{\ast}=\left(  \mathbf{L}_{b}^{\prime}\mathbf{X}_{b}\right)
^{\ast}=\mathbf{X}_{b}^{\ast}\mathbf{L}_{b}^{\prime\ast}
\label{pf.prop.Wls.WI.4}%
\end{equation}
(since the antipode is a $\mathbf{k}$-algebra anti-morphism). But Lemma
\ref{lem.W-S} yields $\mathbf{L}_{b}^{\ast}=\mathbf{L}_{b}$ and $\mathbf{L}%
_{b}^{\prime\ast}=\mathbf{L}_{b}^{\prime}$. Hence, we can rewrite
(\ref{pf.prop.Wls.WI.4}) as
\begin{equation}
\mathbf{L}_{b}=\mathbf{X}_{b}^{\ast}\mathbf{L}_{b}^{\prime}.
\label{pf.prop.Wls.WI.4b}%
\end{equation}


However, $w\in W_{I}$ and $\mathbf{L}_{a}^{\prime},\mathbf{L}_{b}^{\prime}%
\in\mathbf{k}\left[  W_{I}\right]  $ and
\[
\mathbf{X}_{a}=\sum_{x\in R}a^{\ell\left(  x\right)  }%
x\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \mathbf{X}_{b}^{\ast}%
=\sum_{x\in R}b^{\ell\left(  x\right)  }x^{-1}%
\]
(the latter equality follows from $\mathbf{X}_{b}=\sum_{x\in R}b^{\ell\left(
x\right)  }x$ by the definition of the antipode). Hence, Lemma \ref{lem.coset}
(applied to $G=W$ and $H=W_{I}$ and $p_{x}=a^{\ell\left(  x\right)  }$ and
$q_{x}=b^{\ell\left(  x\right)  }$ and $p=\mathbf{X}_{a}$ and $q=\mathbf{X}%
_{b}^{\ast}$ and $\alpha=\mathbf{L}_{a}^{\prime}$ and $\beta=\mathbf{L}%
_{b}^{\prime}$) yields%
\[
\left[  w\right]  \left(  \mathbf{L}_{a}^{\prime}\mathbf{X}_{a}\mathbf{X}%
_{b}^{\ast}\mathbf{L}_{b}^{\prime}\right)  =\left(  \sum_{x\in R}%
a^{\ell\left(  x\right)  }b^{\ell\left(  x\right)  }\right)  \left[  w\right]
\left(  \mathbf{L}_{a}^{\prime}\mathbf{L}_{b}^{\prime}\right)  .
\]
Using (\ref{pf.prop.Wls.WI.3}) and (\ref{pf.prop.Wls.WI.4b}), we can rewrite
this as%
\begin{equation}
\left[  w\right]  \left(  \mathbf{L}_{a}\mathbf{L}_{b}\right)  =\left(
\sum_{x\in R}a^{\ell\left(  x\right)  }b^{\ell\left(  x\right)  }\right)
\left[  w\right]  \left(  \mathbf{L}_{a}^{\prime}\mathbf{L}_{b}^{\prime
}\right)  . \label{pf.prop.Wls.WI.6a}%
\end{equation}
The same argument (with the roles of $a$ and $b$ interchanged) shows that%
\begin{equation}
\left[  w\right]  \left(  \mathbf{L}_{b}\mathbf{L}_{a}\right)  =\left(
\sum_{x\in R}b^{\ell\left(  x\right)  }a^{\ell\left(  x\right)  }\right)
\left[  w\right]  \left(  \mathbf{L}_{b}^{\prime}\mathbf{L}_{a}^{\prime
}\right)  . \label{pf.prop.Wls.WI.6b}%
\end{equation}


Clearly, the factors $\sum_{x\in R}a^{\ell\left(  x\right)  }b^{\ell\left(
x\right)  }$ and $\sum_{x\in R}b^{\ell\left(  x\right)  }a^{\ell\left(
x\right)  }$ on the right-hand sides of the two equalities
(\ref{pf.prop.Wls.WI.6a}) and (\ref{pf.prop.Wls.WI.6b}) are equal, and are not
zero divisors (since the polynomial ring $\mathbb{Z}\left[  a,b\right]  $ is
an integral domain, and since $R$ is nonempty, so that $\sum_{x\in R}%
a^{\ell\left(  x\right)  }b^{\ell\left(  x\right)  }\neq0$ in $\mathbb{Z}%
\left[  a,b\right]  $). Hence, these two equalities show that $\left[
w\right]  \left(  \mathbf{L}_{a}\mathbf{L}_{b}\right)  =\left[  w\right]
\left(  \mathbf{L}_{b}\mathbf{L}_{a}\right)  $ if and only if $\left[
w\right]  \left(  \mathbf{L}_{a}^{\prime}\mathbf{L}_{b}^{\prime}\right)
=\left[  w\right]  \left(  \mathbf{L}_{b}^{\prime}\mathbf{L}_{a}^{\prime
}\right)  $.

In other words, $w$ is $W$-length-symmetric if and only if $w$ is $W_{I}%
$-length-symmetric (since we know that $w$ is $W$-length-symmetric if and only
if $\left[  w\right]  \left(  \mathbf{L}_{a}\mathbf{L}_{b}\right)  =\left[
w\right]  \left(  \mathbf{L}_{b}\mathbf{L}_{a}\right)  $, whereas $w$ is
$W_{I}$-length-symmetric if and only if $\left[  w\right]  \left(
\mathbf{L}_{a}^{\prime}\mathbf{L}_{b}^{\prime}\right)  =\left[  w\right]
\left(  \mathbf{L}_{b}^{\prime}\mathbf{L}_{a}^{\prime}\right)  $). This proves
Proposition \ref{prop.Wls.WI}.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.Wls.len2}.]From $\ell\left(  w\right)  \leq2$,
we know that we can write $w$ as a product of at most two simple reflections.
Let $I$ be the set of all simple reflections appearing in this product. Then,
$w\in W_{I}$ and $\left\vert I\right\vert \leq2$. Hence, Corollary
\ref{cor.Wls.dih} (applied to the Coxeter system $\left(  W_{I},I\right)  $
instead of $\left(  W,S\right)  $) yields that $w$ is $W_{I}$%
-length-symmetric. By Proposition \ref{prop.Wls.WI}, this entails that $w$ is
$W$-length-symmetric. Thus, Proposition \ref{prop.Wls.len2} is proved.
\end{proof}

\subsection{Orthogonal decompositions}

\begin{proof}
[Proof of Proposition \ref{prop.Wls.I+J}.]Let $\mathbf{k}$ be any commutative
ring, and $a,b\in\mathbf{k}$.

Since $S=I\oplus J$, we have $W\cong W_{I}\times W_{J}$; more specifically,
the map%
\begin{align}
W_{I}\times W_{J}  &  \rightarrow W,\nonumber\\
\left(  x,y\right)   &  \mapsto xy \label{pf.prop.Wls.I+J.bijgp}%
\end{align}
is a group isomorphism. This isomorphism entails that the $\mathbf{k}$-linear
map%
\begin{align}
\Phi:\mathbf{k}\left[  W_{I}\right]  \otimes\mathbf{k}\left[  W_{J}\right]
&  \rightarrow\mathbf{k}\left[  W\right]  ,\nonumber\\
x\otimes y  &  \mapsto xy \label{pf.prop.Wls.I+J.bijal}%
\end{align}
is a $\mathbf{k}$-algebra isomorphism (because there is a $\mathbf{k}$-algebra
isomorphism%
\begin{align*}
\mathbf{k}\left[  W_{I}\times W_{J}\right]   &  \rightarrow\mathbf{k}\left[
W_{I}\right]  \otimes\mathbf{k}\left[  W_{J}\right]  ,\\
\left(  x,y\right)   &  \mapsto x\otimes y,
\end{align*}
and if we identify $\mathbf{k}\left[  W_{I}\right]  \otimes\mathbf{k}\left[
W_{J}\right]  $ with $\mathbf{k}\left[  W_{I}\times W_{J}\right]  $ along this
isomorphism, then (\ref{pf.prop.Wls.I+J.bijal}) is the linearization of the
group isomorphism (\ref{pf.prop.Wls.I+J.bijgp})). The construction of $\Phi$
shows that any $\alpha\in\mathbf{k}\left[  W_{I}\right]  $ and $\beta
\in\mathbf{k}\left[  W_{J}\right]  $ satisfy%
\begin{equation}
\left[  uv\right]  \left(  \Phi\left(  \alpha\otimes\beta\right)  \right)
=\left[  u\right]  \left(  \alpha\right)  \cdot\left[  v\right]  \left(
\beta\right)  . \label{pf.prop.Wls.I+J.uvPhi}%
\end{equation}
(Indeed, this equality is $\mathbf{k}$-linear in $\alpha$ and $\beta$, and
thus needs only be checked for all $\alpha\in W_{I}$ and $\beta\in W_{J}$; but
then it boils down to \textquotedblleft$\alpha\beta=uv$ if and only if
$\alpha=u$ and $\beta=v$\textquotedblright.)

Moreover, any element of $W_{I}$ commutes with any element of $W_{J}$; thus,
for any $x\in W_{I}$ and $y\in W_{J}$, we have%
\begin{equation}
\ell\left(  xy\right)  =\ell\left(  x\right)  +\ell\left(  y\right)
\label{pf.prop.Wls.I+J.l=l+l}%
\end{equation}
(since any reduced expression for $xy$ can be transformed into a reduced
expression for $x$ times a reduced expression for $y$ by commuting its
$I$-factors to the left of all its $J$-factors\footnote{In more detail: Let
$x\in W_{I}$ and $y\in W_{J}$. Let $s_{1}s_{2}\cdots s_{p}$ be any reduced
expression for $xy$. Then, $p=\ell\left(  xy\right)  $. Moreover, each $s_{i}$
belongs to $S=I\sqcup J$, hence lies in either $I$ or $J$. Since each $i\in I$
commutes with each $j\in J$ (because $S=I\oplus J$ is an orthogonal
decomposition), we can freely move all the $s_{i}$ that belong to $I$ in the
reduced expression $s_{1}s_{2}\cdots s_{p}$ past the $s_{i}$ that belong to
$J$. In particular, we can move all the former $s_{i}$ to the left end of the
reduced expression, while letting the latter $s_{i}$ accumulate on the right
end. As a consequence, our reduced expression assumes the form $t_{1}%
t_{2}\cdots t_{p}$, where the first $k$ factors $t_{1},t_{2},\ldots,t_{k}$
(for some $k\in\left\{  0,1,\ldots,p\right\}  $) belong to $I$ while the
remaining $p-k$ factors $t_{k+1},t_{k+2},\ldots,t_{p}$ belong to $J$. Thus, of
course, $t_{1}t_{2}\cdots t_{k}\in W_{I}$ and $t_{k+1}t_{k+2}\cdots t_{p}\in
W_{J}$. Hence,%
\[
xy=s_{1}s_{2}\cdots s_{p}=t_{1}t_{2}\cdots t_{p}=\underbrace{\left(
t_{1}t_{2}\cdots t_{k}\right)  }_{\in W_{I}}\underbrace{\left(  t_{k+1}%
t_{k+2}\cdots t_{p}\right)  }_{\in W_{J}}.
\]
In other words, the map (\ref{pf.prop.Wls.I+J.bijgp}) sends the pairs $\left(
x,y\right)  \in W_{I}\times W_{J}$ and $\left(  t_{1}t_{2}\cdots
t_{k},\ t_{k+1}t_{k+2}\cdots t_{p}\right)  \in W_{I}\times W_{J}$ to the same
image. Since this map is injective (being a group isomorphism), it thus
follows that these pairs are equal. That is, $x=t_{1}t_{2}\cdots t_{k}$ and
$y=t_{k+1}t_{k+2}\cdots t_{p}$. Hence, $\ell\left(  x\right)  \leq k$ and
$\ell\left(  y\right)  \leq p-k$. Summing these two inequalities, we obtain
$\ell\left(  x\right)  +\ell\left(  y\right)  \leq k+\left(  p-k\right)
=p=\ell\left(  xy\right)  $. Combining this with $\ell\left(  xy\right)
\leq\ell\left(  x\right)  +\ell\left(  y\right)  $ (which is a general
property of Coxeter lengths in any Coxeter group), we obtain $\ell\left(
xy\right)  =\ell\left(  x\right)  +\ell\left(  y\right)  $. This proves
(\ref{pf.prop.Wls.I+J.l=l+l}).}). But Definition \ref{def.Lq} yields%
\[
\mathbf{L}_{a}\left(  W_{I}\right)  =\sum_{x\in W_{I}}a^{\ell\left(  x\right)
}x\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \mathbf{L}_{a}\left(
W_{J}\right)  =\sum_{y\in W_{J}}a^{\ell\left(  y\right)  }y,
\]
so that%
\begin{align*}
\mathbf{L}_{a}\left(  W_{I}\right)  \otimes\mathbf{L}_{a}\left(  W_{J}\right)
&  =\left(  \sum_{x\in W_{I}}a^{\ell\left(  x\right)  }x\right)
\otimes\left(  \sum_{y\in W_{J}}a^{\ell\left(  y\right)  }y\right) \\
&  =\sum_{\left(  x,y\right)  \in W_{I}\times W_{J}}\ \ \underbrace{a^{\ell
\left(  x\right)  }a^{\ell\left(  y\right)  }}_{\substack{=a^{\ell\left(
x\right)  +\ell\left(  y\right)  }=a^{\ell\left(  xy\right)  }\\\text{(by
(\ref{pf.prop.Wls.I+J.l=l+l}))}}}x\otimes y=\sum_{\left(  x,y\right)  \in
W_{I}\times W_{J}}a^{\ell\left(  xy\right)  }x\otimes y.
\end{align*}
Applying the linear map $\Phi$ to this equality, we obtain%
\begin{align*}
\Phi\left(  \mathbf{L}_{a}\left(  W_{I}\right)  \otimes\mathbf{L}_{a}\left(
W_{J}\right)  \right)   &  =\sum_{\left(  x,y\right)  \in W_{I}\times W_{J}%
}a^{\ell\left(  xy\right)  }\underbrace{\Phi\left(  x\otimes y\right)  }%
_{=xy}=\sum_{\left(  x,y\right)  \in W_{I}\times W_{J}}a^{\ell\left(
xy\right)  }xy\\
&  =\sum_{w\in W}a^{\ell\left(  w\right)  }w\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }w\text{ for }xy\text{,}\\
\text{since the map (\ref{pf.prop.Wls.I+J.bijgp}) is a bijection}%
\end{array}
\right) \\
&  =\mathbf{L}_{a}\left(  W\right)  .
\end{align*}
Similarly, $\Phi\left(  \mathbf{L}_{b}\left(  W_{I}\right)  \otimes
\mathbf{L}_{b}\left(  W_{J}\right)  \right)  =\mathbf{L}_{b}\left(  W\right)
$. Since $\Phi$ is an algebra morphism, we now have%
\begin{align*}
&  \Phi\left(  \left(  \mathbf{L}_{a}\left(  W_{I}\right)  \otimes
\mathbf{L}_{a}\left(  W_{J}\right)  \right)  \left(  \mathbf{L}_{b}\left(
W_{I}\right)  \otimes\mathbf{L}_{b}\left(  W_{J}\right)  \right)  \right) \\
&  =\underbrace{\Phi\left(  \mathbf{L}_{a}\left(  W_{I}\right)  \otimes
\mathbf{L}_{a}\left(  W_{J}\right)  \right)  }_{=\mathbf{L}_{a}\left(
W\right)  }\cdot\underbrace{\Phi\left(  \mathbf{L}_{b}\left(  W_{I}\right)
\otimes\mathbf{L}_{b}\left(  W_{J}\right)  \right)  }_{=\mathbf{L}_{b}\left(
W\right)  }\\
&  =\mathbf{L}_{a}\left(  W\right)  \cdot\mathbf{L}_{b}\left(  W\right)  ,
\end{align*}
so that%
\begin{align*}
&  \mathbf{L}_{a}\left(  W\right)  \cdot\mathbf{L}_{b}\left(  W\right) \\
&  =\Phi\left(  \underbrace{\left(  \mathbf{L}_{a}\left(  W_{I}\right)
\otimes\mathbf{L}_{a}\left(  W_{J}\right)  \right)  \left(  \mathbf{L}%
_{b}\left(  W_{I}\right)  \otimes\mathbf{L}_{b}\left(  W_{J}\right)  \right)
}_{=\mathbf{L}_{a}\left(  W_{I}\right)  \cdot\mathbf{L}_{b}\left(
W_{I}\right)  \otimes\mathbf{L}_{a}\left(  W_{J}\right)  \cdot\mathbf{L}%
_{b}\left(  W_{J}\right)  }\right) \\
&  =\Phi\left(  \mathbf{L}_{a}\left(  W_{I}\right)  \cdot\mathbf{L}_{b}\left(
W_{I}\right)  \otimes\mathbf{L}_{a}\left(  W_{J}\right)  \cdot\mathbf{L}%
_{b}\left(  W_{J}\right)  \right)  .
\end{align*}
Thus,%
\begin{align}
&  \left[  uv\right]  \left(  \mathbf{L}_{a}\left(  W\right)  \cdot
\mathbf{L}_{b}\left(  W\right)  \right) \nonumber\\
&  =\left[  uv\right]  \left(  \Phi\left(  \mathbf{L}_{a}\left(  W_{I}\right)
\cdot\mathbf{L}_{b}\left(  W_{I}\right)  \otimes\mathbf{L}_{a}\left(
W_{J}\right)  \cdot\mathbf{L}_{b}\left(  W_{J}\right)  \right)  \right)
\nonumber\\
&  =\left[  u\right]  \left(  \mathbf{L}_{a}\left(  W_{I}\right)
\cdot\mathbf{L}_{b}\left(  W_{I}\right)  \right)  \cdot\left[  v\right]
\left(  \mathbf{L}_{a}\left(  W_{J}\right)  \cdot\mathbf{L}_{b}\left(
W_{J}\right)  \right)  \label{pf.prop.Wls.I+J.6a}%
\end{align}
(by (\ref{pf.prop.Wls.I+J.uvPhi}), applied to $\alpha=\mathbf{L}_{a}\left(
W_{I}\right)  \cdot\mathbf{L}_{b}\left(  W_{I}\right)  $ and $\beta
=\mathbf{L}_{a}\left(  W_{J}\right)  \cdot\mathbf{L}_{b}\left(  W_{J}\right)
$). The same argument (with the roles of $a$ and $b$ interchanged) shows that%
\begin{align}
&  \left[  uv\right]  \left(  \mathbf{L}_{b}\left(  W\right)  \cdot
\mathbf{L}_{a}\left(  W\right)  \right) \nonumber\\
&  =\left[  u\right]  \left(  \mathbf{L}_{b}\left(  W_{I}\right)
\cdot\mathbf{L}_{a}\left(  W_{I}\right)  \right)  \cdot\left[  v\right]
\left(  \mathbf{L}_{b}\left(  W_{J}\right)  \cdot\mathbf{L}_{a}\left(
W_{J}\right)  \right)  . \label{pf.prop.Wls.I+J.6b}%
\end{align}


However, $\left[  u\right]  \left(  \mathbf{L}_{a}\left(  W_{I}\right)
\cdot\mathbf{L}_{b}\left(  W_{I}\right)  \right)  =\left[  u\right]  \left(
\mathbf{L}_{b}\left(  W_{I}\right)  \cdot\mathbf{L}_{a}\left(  W_{I}\right)
\right)  $ (since $u$ is $W_{I}$-length-symmetric) and $\left[  v\right]
\left(  \mathbf{L}_{a}\left(  W_{J}\right)  \cdot\mathbf{L}_{b}\left(
W_{J}\right)  \right)  =\left[  v\right]  \left(  \mathbf{L}_{b}\left(
W_{J}\right)  \cdot\mathbf{L}_{a}\left(  W_{J}\right)  \right)  $ (since $v$
is $W_{J}$-length-symmetric). Thus, the right-hand sides of
(\ref{pf.prop.Wls.I+J.6a}) and (\ref{pf.prop.Wls.I+J.6b}) are equal (factor by
factor). Consequently, so are the left-hand sides. In other words,
\[
\left[  uv\right]  \left(  \mathbf{L}_{a}\left(  W\right)  \cdot\mathbf{L}%
_{b}\left(  W\right)  \right)  =\left[  uv\right]  \left(  \mathbf{L}%
_{b}\left(  W\right)  \cdot\mathbf{L}_{a}\left(  W\right)  \right)  .
\]
This shows that $uv$ is $W$-length-symmetric. This proves Proposition
\ref{prop.Wls.I+J}.
\end{proof}

\subsection{The longest word}

\begin{lemma}
\label{lem.w0Lq}Let $\left(  W,S\right)  $ be a Coxeter system, where $W$ is
finite. Let $q\in\mathbf{k}$ be invertible. Then,%
\[
w_{0}\mathbf{L}_{q}=\mathbf{L}_{q}w_{0}=q^{\ell\left(  w_{0}\right)
}\mathbf{L}_{1/q}.
\]

\end{lemma}

\begin{proof}
As we know, $w_{0}$ is an involution; thus, $w_{0}w_{0}=1$ and $w_{0}%
^{-1}=w_{0}$. Moreover,%
\begin{equation}
\ell\left(  w_{0}w\right)  =\ell\left(  w_{0}\right)  -\ell\left(  w\right)
\label{pf.lem.w0Lq.1}%
\end{equation}
for any $w\in W$ (see, e.g., \cite[\S 11.2 (b)]{Lusztig} for a proof). But the
definition of $\mathbf{L}_{1/q}$ yields
\begin{equation}
\mathbf{L}_{1/q}=\sum_{w\in W}\left(  1/q\right)  ^{\ell\left(  w\right)  }w.
\label{pf.lem.w0Lq.2}%
\end{equation}
Now, the definition of $\mathbf{L}_{q}$ yields%
\begin{align*}
w_{0}\mathbf{L}_{q}  &  =w_{0}\sum_{w\in W}q^{\ell\left(  w\right)  }%
w=\sum_{w\in W}q^{\ell\left(  w\right)  }w_{0}w\\
&  =\sum_{w\in W}\underbrace{q^{\ell\left(  w_{0}w\right)  }}%
_{\substack{=q^{\ell\left(  w_{0}\right)  -\ell\left(  w\right)  }\\\text{(by
(\ref{pf.lem.w0Lq.1}))}}}\underbrace{w_{0}w_{0}}_{=1}%
w\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }w_{0}w\\
\text{for }w\text{ in the sum}%
\end{array}
\right) \\
&  =\sum_{w\in W}\underbrace{q^{\ell\left(  w_{0}\right)  -\ell\left(
w\right)  }}_{\substack{=q^{\ell\left(  w_{0}\right)  }/q^{\ell\left(
w\right)  }\\=q^{\ell\left(  w_{0}\right)  }\left(  1/q\right)  ^{\ell\left(
w\right)  }}}w=q^{\ell\left(  w_{0}\right)  }\underbrace{\sum_{w\in W}\left(
1/q\right)  ^{\ell\left(  w\right)  }w}_{\substack{=\mathbf{L}_{1/q}%
\\\text{(by (\ref{pf.lem.w0Lq.2}))}}}=q^{\ell\left(  w_{0}\right)  }%
\mathbf{L}_{1/q}.
\end{align*}
Applying the antipode to both sides of this equality, we obtain%
\[
\left(  w_{0}\mathbf{L}_{q}\right)  ^{\ast}=q^{\ell\left(  w_{0}\right)
}\mathbf{L}_{1/q}^{\ast}.
\]
Since $\left(  w_{0}\mathbf{L}_{q}\right)  ^{\ast}=\mathbf{L}_{q}^{\ast}%
w_{0}^{\ast}$ (because the antipode is an algebra anti-morphism), we can
rewrite this as%
\[
\mathbf{L}_{q}^{\ast}w_{0}^{\ast}=q^{\ell\left(  w_{0}\right)  }%
\mathbf{L}_{1/q}^{\ast}.
\]
Using Lemma \ref{lem.W-S} and the obvious fact that $w_{0}^{\ast}=w_{0}%
^{-1}=w_{0}$, we can rewrite this as%
\[
\mathbf{L}_{q}w_{0}=q^{\ell\left(  w_{0}\right)  }\mathbf{L}_{1/q}.
\]
Combining this with $w_{0}\mathbf{L}_{q}=q^{\ell\left(  w_{0}\right)
}\mathbf{L}_{1/q}$, we obtain the whole claim of Lemma \ref{lem.w0Lq}.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.Wls.ref}.]Recall that $w_{0}$ is an
involution, so that $w_{0}=w_{0}^{-1}$ and $w_{0}w_{0}=1$. Hence,
$w_{0}\left(  w_{0}w\right)  =\underbrace{w_{0}w_{0}}_{=1}w=w$. Thus, the
relation between $w$ and $w_{0}w$ is mutual. Hence, we only need to prove the
\textquotedblleft only if\textquotedblright\ direction of Proposition
\ref{prop.Wls.ref} (as the \textquotedblleft if\textquotedblright\ direction
will then follow by applying the \textquotedblleft only if\textquotedblright%
\ direction to $w_{0}w$ instead of $w$).

So let us assume that $w$ is $W$-length-symmetric. We must show that $w_{0}w$
is $W$-length-symmetric. Set $u:=w_{0}w$. Thus, we must show that $u$ is $W$-length-symmetric.

Fix a commutative ring $\mathbf{k}$ and $a,b\in\mathbf{k}$. We must show that
$\left[  u\right]  \left(  \mathbf{L}_{a}\mathbf{L}_{b}\right)  =\left[
u\right]  \left(  \mathbf{L}_{b}\mathbf{L}_{a}\right)  $. We WLOG assume that
$\mathbf{k}$ is a ring of Laurent polynomials and $a$ and $b$ are two
indeterminates (since the claim we are proving is a polynomial identity in $a$
and $b$). Thus, $a$ and $b$ are invertible in $\mathbf{k}$. Since $w$ is
$W$-length-symmetric, we have $\left[  w\right]  \left(  \mathbf{L}%
_{a}\mathbf{L}_{b}\right)  =\left[  w\right]  \left(  \mathbf{L}_{b}%
\mathbf{L}_{a}\right)  $ and likewise
\[
\left[  w\right]  \left(  \mathbf{L}_{a}\mathbf{L}_{1/b}\right)  =\left[
w\right]  \left(  \mathbf{L}_{1/b}\mathbf{L}_{a}\right)  .
\]


From Lemma \ref{lem.w0Lq}, we have%
\[
w_{0}\mathbf{L}_{a}=\mathbf{L}_{a}w_{0}=a^{\ell\left(  w_{0}\right)
}\mathbf{L}_{1/a}%
\]
and%
\[
w_{0}\mathbf{L}_{b}=\mathbf{L}_{b}w_{0}=b^{\ell\left(  w_{0}\right)
}\mathbf{L}_{1/b}.
\]


From $u=w_{0}w=w_{0}^{-1}w$ (since $w_{0}=w_{0}^{-1}$), we obtain%
\[
\left[  u\right]  \left(  \mathbf{L}_{a}\mathbf{L}_{b}\right)  =\left[
w_{0}^{-1}w\right]  \left(  \mathbf{L}_{a}\mathbf{L}_{b}\right)  =\left[
w\right]  \left(  w_{0}\mathbf{L}_{a}\mathbf{L}_{b}\right)
\]
(since Lemma \ref{lem.coeff} (applied to $w_{0}$, $1$ and $\mathbf{L}%
_{a}\mathbf{L}_{b}$ instead of $g$, $h$ and $a$) yields $\left[  w\right]
\left(  w_{0}\mathbf{L}_{a}\mathbf{L}_{b}\right)  =\left[  w_{0}^{-1}w\right]
\left(  \mathbf{L}_{a}\mathbf{L}_{b}\right)  $). Therefore,%
\[
\left[  u\right]  \left(  \mathbf{L}_{a}\mathbf{L}_{b}\right)  =\left[
w\right]  \left(  \underbrace{w_{0}\mathbf{L}_{a}}_{=\mathbf{L}_{a}w_{0}%
}\mathbf{L}_{b}\right)  =\left[  w\right]  \left(  \mathbf{L}_{a}%
w_{0}\mathbf{L}_{b}\right)  .
\]
The same argument (with the roles of $a$ and $b$ swapped) shows that%
\[
\left[  u\right]  \left(  \mathbf{L}_{b}\mathbf{L}_{a}\right)  =\left[
w\right]  \left(  \mathbf{L}_{b}w_{0}\mathbf{L}_{a}\right)  .
\]
The right-hand sides of these two equalities are equal, since%
\begin{align*}
\left[  w\right]  \left(  \mathbf{L}_{a}\underbrace{w_{0}\mathbf{L}_{b}%
}_{=b^{\ell\left(  w_{0}\right)  }\mathbf{L}_{1/b}}\right)   &  =b^{\ell
\left(  w_{0}\right)  }\underbrace{\left[  w\right]  \left(  \mathbf{L}%
_{a}\mathbf{L}_{1/b}\right)  }_{=\left[  w\right]  \left(  \mathbf{L}%
_{1/b}\mathbf{L}_{a}\right)  }=b^{\ell\left(  w_{0}\right)  }\left[  w\right]
\left(  \mathbf{L}_{1/b}\mathbf{L}_{a}\right) \\
&  =\left[  w\right]  \left(  \underbrace{b^{\ell\left(  w_{0}\right)
}\mathbf{L}_{1/b}}_{=\mathbf{L}_{b}w_{0}}\mathbf{L}_{a}\right)  =\left[
w\right]  \left(  \mathbf{L}_{b}w_{0}\mathbf{L}_{a}\right)  .
\end{align*}
Hence, so are the left-hand sides. That is, $\left[  u\right]  \left(
\mathbf{L}_{a}\mathbf{L}_{b}\right)  =\left[  u\right]  \left(  \mathbf{L}%
_{b}\mathbf{L}_{a}\right)  $. This shows that $u$ is $W$-length-symmetric.
This proves the \textquotedblleft only if\textquotedblright\ direction of
Proposition \ref{prop.Wls.ref}, and thus (as explained above) the whole proposition.
\end{proof}

\section{Application: Separable permutations (TODO)}

We now apply the above results to prove a theorem we promised in the introduction:

\begin{theorem}
\label{thm.Sn.sep-len-sym}Let $n\in\mathbb{N}$. Consider the symmetric group
$S_{n}$ as a Coxeter group, as in Example \ref{exa.LqSn}. A permutation $w\in
S_{n}$ is said to be \emph{separable} if and only if there exist no four
elements $a<b<c<d$ of $\left\{  1,2,\ldots,n\right\}  $ satisfying%
\[
\left(  w\left(  b\right)  <w\left(  d\right)  <w\left(  a\right)  <w\left(
c\right)  \ \ \ \ \ \ \ \ \ \ \text{or}\ \ \ \ \ \ \ \ \ \ w\left(  c\right)
<w\left(  a\right)  <w\left(  d\right)  <w\left(  b\right)  \right)  .
\]
Then, each separable permutation $w\in S_{n}$ is $S_{n}$-length-symmetric.
\end{theorem}

\begin{proof}
[Proof sketch.]Among the many characterizations of separable permutations
given in \cite[\textquotedblleft From a pruned tree to a polynomial
interchange and a separable permutation\textquotedblright, Theorem]{Ghys16},
we will need the fourth one: A permutation $w\in S_{n}$ is separable if and
only if it can be obtained from several copies of the trivial permutation on
one object (i.e., the permutation $\operatorname*{id}\in S_{1}$) by successive
$\oplus$ and $\ominus$ operations. Here, the operation $\oplus$ sends two
permutations $u\in S_{k}$ and $v\in S_{\ell}$ to the permutation $u\oplus v\in
S_{k+\ell}$ that acts as $u$ on the numbers $1,2,\ldots,k$ and as (a shifted
copy of) $v$ on the numbers $k+1,k+2,\ldots,k+\ell$, whereas the operation
$\ominus$ sends two permutations $u\in S_{k}$ and $v\in S_{\ell}$ to the
permutation $\left(  \left(  uw_{0,k}\right)  \oplus\left(  vw_{0,\ell
}\right)  \right)  w_{0,k+\ell}\in S_{k+\ell}$, where $w_{0,m}$ denotes the
longest word of $S_{m}$. This recursive definition allows for proving the
claim by induction, using Proposition \ref{prop.Wls.I+J} (to show that
$u\oplus v$ is $S_{n}$-length-symmetric whenever $u$ and $v$ are) and
Proposition \ref{prop.Wls.ref} and Proposition \ref{prop.Wls.inverse} (to show
that $ww_{0}$ is $S_{n}$-length-symmetric whenever $w$ is). We omit the details.

TODO: give the details...
\end{proof}

\bigskip

\begin{thebibliography}{999999999}                                                                                        %


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\bibitem[BjoBre05]{BjoBre}\href{https://doi.org/10.1007/3-540-27596-7}{Anders
Bjorner, Francesco Brenti, \textit{Combinatorics of Coxeter Groups}, Springer
2005}.\newline See \url{https://www.mat.uniroma2.it/~brenti/correct.ps} for errata.

\bibitem[Bourba02]{Bourbaki-Lie4-6}Nicolas Bourbaki, \textit{Lie Groups and
Lie Algebras, Chapters 4--6}, Springer 2002.

\bibitem[Curtis85]{Curtis}%
\href{https://doi.org/10.1016/0021-8693(85)90125-5}{Charles W. Curtis,
\textit{On Lusztig's isomorphism theorem for Hecke algebras}, Journal of
Algebra \textbf{92} (1985), Issue 2, pp. 348--365.}

\bibitem[Davis25]{Davis}%
\href{https://people.math.osu.edu/davis.12/second_edition.pdf}{Michael W.
Davis, \textit{The Geometry and Topology of Coxeter Groups}, 2nd edition
2025.}\newline See
\url{https://www.cip.ifi.lmu.de/~grinberg/algebra/davis-gtcox-errata.pdf} for
unofficial errata.

\bibitem[Digne08]{Digne}Fran\c{c}ois Digne, \textit{Alg\`{e}bres de Hecke}, 5
December 2008.\newline\url{https://www.lamfa.u-picardie.fr/digne/hecke.pdf}

\bibitem[GaeGao20a]{GaeGao20a}%
\href{https://arxiv.org/abs/1905.09331v2}{Christian Gaetz, Yibo Gao,
\textit{Separable elements in Weyl groups}, arXiv:1905.09331v2, Adv. Appl.
Math. \textbf{113} (2020).}

\bibitem[GaeGao20b]{GaeGao20b}%
\href{https://arxiv.org/abs/1911.11172v2}{Christian Gaetz, Yibo Gao,
\textit{Separable elements and splittings of Weyl groups}, arXiv:1911.11172v2,
Adv. Math. \textbf{374} (2020).}

\bibitem[Gaetz21]{Gaetz21}%
\href{https://hdl.handle.net/1721.1/139101}{Christian Gaetz, \textit{New
combinatorics of the weak and strong Bruhat orders}, PhD thesis at MIT, 2021.}

\bibitem[Ghys16]{Ghys16}\href{https://arxiv.org/abs/1612.06373v4}{\'{E}tienne
Ghys, \textit{A Singular Mathematical Promenade}, arXiv:1612.06373v4}.\newline
Published in: ENS \'{e}ditions, Lyon, 2017.

\bibitem[Hiller82]{Hiller}Howard Hiller, \textit{Geometry of Coxeter groups},
Pitman Advanced Publishing Program 1982.

\bibitem[Humphr90]{Humphreys}James E. Humphreys, \textit{Reflection groups and
Coxeter groups}, Cambridge University Press 1990.

\bibitem[Korotk16]{Korotk16}%
\href{https://doi.org/10.1007/s10958-017-3413-5}{S. Korotkikh, \textit{The
Mallows measures on the hyperoctahedral group}, Zapiski Nauchnykh Seminarov
POMI \textbf{448} (2016), pp. 151--164.}

\bibitem[Luszti14]{Lusztig}\href{http://arxiv.org/abs/math/0208154v2}{George
Lusztig, \textit{Hecke algebras with unequal parameters},
arXiv:math/0208154v2.}

\bibitem[Heckma18]{Heckman}Gert Heckman, \textit{Coxeter groups}, 10 October
2018.\newline\url{https://www.math.ru.nl/~heckman/CoxeterGroups.pdf}

\bibitem[Sam21]{Sam21}Steven V. Sam, \textit{Notes for Math 264C (Spring
2021)}, 8 May 2021.\newline\url{https://mathweb.ucsd.edu/~ssam/old/21S-264C/notes-264C.pdf}

\bibitem[Schrem23]{Schremmer}Felix Schremmer, \textit{Topics in Algebra II:
Coxeter groups}, 26 April 2023.\newline\url{https://www.math.cuhk.edu.hk/course_builder/2223/math6032/lecture-notes-coxeter.pdf}

\bibitem[Werner12]{Werner}Annette Werner, \textit{Geb\"{a}ude}, Sommersemester
2012, 27 August 2012.\newline%
\url{https://www.uni-frankfurt.de/115698695/gebaeude.pdf}\newline See
\url{https://www.uni-frankfurt.de/115681659/Prof__Dr__Annette_Werner#a_5d3e5bbe-70282236}
for other notes.
\end{thebibliography}


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